∫ WBCHSE/ISC/CBSE - All Important Integrations
90+ Integrals Marks · 2M 3M 4M
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📘 2 Mark Integrals
30+ problemsSolution:
\(x^{2} - 1 = 0\)
\(\Rightarrow x = \pm 1\)
\(|x^{2} - 1| = \begin{cases} + (x^{2} - 1) \text{ when } 1 \le x \le 2 \\ - (x^{2} - 1) \text{ when } 0 \le x < 1 \end{cases}\)
So, \(\; I = \int_{0}^{2} |x^{2} - 1| dx\)
\(= \int_{0}^{1} - (x^{2} - 1) dx + \int_{1}^{2} (x^{2} - 1) dx\)
\(= -\frac{1}{3} [x^{3}]_{0}^{1} + [x]_{0}^{1} + \frac{1}{3} [x^{3}]_{1}^{2} - [x]_{1}^{2}\)
\(= -\frac{1}{3} (1 - 0) + (1 - 0) + \frac{1}{3} (8 - 1) - (2 - 1)\)
\(= -\frac{1}{3} + 1 + \frac{7}{3} - 1\)
\(= 2\)
Hence, \(\int_{0}^{2} |x^{2} - 1| dx = 2\) (Answer)
Solution:
Let \(\; e^{x} - 1 = z^{2} \;\Rightarrow\; e^{x} = z^{2} + 1\)
Differentiating w.r.t. x, we get
\(e^{x} dx = 2z dz\)
\(\Rightarrow (z^{2} + 1) dx = 2z dz\)
\(\therefore dx = \frac{2z}{z^{2} + 1} dz\)
So, \(\; I = \int \frac{dx}{\sqrt{e^{x}-1}}\)
\(= \int \frac{2z}{(z^{2}+1)z} dz\)
\(= 2 \int \frac{dz}{1 + z^{2}}\)
\(= 2 \tan^{-1}(z) + C\)
\(= 2 \tan^{-1}(\sqrt{e^{x}-1}) + C\)
\(\therefore \int \frac{dx}{\sqrt{e^{x}-1}} = 2 \tan^{-1}(\sqrt{e^{x}-1}) + C \quad (Answer)\)
Solution:
Let \(\tan x = z\)
Differentiating w.r.t. x, we get
\(\sec^{2} x \, dx = dz\)
Now, \(\; I = \int \frac{dz}{(\sqrt{3})^{2} + z^{2}}\)
\(= \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{z}{\sqrt{3}} \right) + C\)
\(= \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{\tan x}{\sqrt{3}} \right) + C\)
Hence, \(\displaystyle \int \frac{\sec^{2} x \, dx}{3 + \tan^{2} x} = \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{\tan x}{\sqrt{3}} \right) + C \quad (Answer)\)
Solution:
\(5 - 8x - x^{2}\)
\(= - (x^{2} + 2 \cdot x \cdot 4 + 4^{2}) + 5 + 16\)
\(= (\sqrt{21})^{2} - (x + 4)^{2}\)
So, \(\; I = \int \frac{dx}{5 - 8x - x^{2}}\)
\(= \int \frac{dx}{(\sqrt{21})^{2} - (x + 4)^{2}}\)
\(= \frac{1}{2\sqrt{21}} \log \left| \frac{\sqrt{21} + x + 4}{\sqrt{21} - x - 4} \right| + C\)
Hence, \(\displaystyle \int \frac{dx}{5 - 8x - x^{2}} = \frac{1}{2\sqrt{21}} \log \left| \frac{x + 4 + \sqrt{21}}{\sqrt{21} - x - 4} \right| + C \quad (Answer)\)
Solution:
\(\displaystyle \int \frac{e^{3x} + e^{5x}}{e^{x} + e^{-x}} \, dx\)
\(= \int \frac{e^{3x}(e^{2x} + 1)}{e^{-x}(e^{2x} + 1)} \, dx \qquad [e^{2x} + 1 \ne 0]\)
\(= \int e^{4x} \, dx\)
\(= \frac{e^{4x}}{4} + C \quad (Ans)\)
Solution:
\(\displaystyle \int \frac{\sec x \, dx}{\sec x + \tan x}\)
\(= \int \frac{\sec x (\sec x - \tan x)}{\sec^{2} x - \tan^{2} x} \, dx\)
\(= \int \sec^{2} x \, dx - \int \sec x \tan x \, dx\)
\(= \tan x - \sec x + C \quad (Ans)\)
Solution:
\(I = \int \frac{dx}{1 + \sin x}\)
\(= \int \frac{1 - \sin x}{1 - \sin^{2} x} \, dx\)
\(= \int \frac{1 - \sin x}{\cos^{2} x} \, dx\)
\(= \int \sec^{2} x \, dx - \int \sec x \tan x \, dx\)
\(= \tan x - \sec x + C \quad (Ans)\)
Solution:
\(I = \int \frac{dx}{1 - \cos x}\)
\(= \int \frac{1 + \cos x}{1 - \cos^{2} x} \, dx\)
\(= \int \frac{1 + \cos x}{\sin^{2} x} \, dx\)
\(= \int \text{cosec}^{2} x \, dx + \int \text{cosec} x \cot x \, dx\)
\(= - \cot x - \text{cosec} x + C \quad (Ans)\)
Solution:
\(I = \int \frac{dx}{\sqrt{ax+b} + \sqrt{ax-b}}\)
\(= \int \frac{\sqrt{ax+b} - \sqrt{ax-b}}{ax+b - (ax-b)} \, dx\)
\(= \int \frac{\sqrt{ax+b}}{2b} \, dx - \int \frac{\sqrt{ax-b}}{2b} \, dx\)
\(= \frac{1}{2b} \int (ax+b)^{1/2} dx - \frac{1}{2b} \int (ax-b)^{1/2} dx\)
\(= \frac{1}{2b} \cdot \frac{(ax+b)^{3/2}}{\frac{3}{2}a} - \frac{1}{2b} \cdot \frac{(ax-b)^{3/2}}{\frac{3}{2}a} + C\)
\(= \frac{1}{3ab} (ax+b)^{3/2} - \frac{1}{3ab} (ax-b)^{3/2} + C\)
\(= \frac{1}{3ab} \left[ (ax+b)\sqrt{ax+b} - (ax-b)\sqrt{ax-b} \right] + C \quad (Ans)\)
Solution:
\(I = \int \frac{x^{5}}{\sqrt{1 + x^{3}}} \, dx\)
\(= \frac{1}{3} \int \frac{3x^{2} \cdot x^{3}}{\sqrt{1 + x^{3}}} \, dx\)
Let, \(\; x^{3} = z\)
\(3x^{2} dx = dz\)
\(I = \frac{1}{3} \int \frac{z \, dz}{\sqrt{1 + z}}\)
\(\Rightarrow I = \frac{1}{3} \int \frac{(z+1 - 1)}{\sqrt{z+1}} \, dz\)
\(I = \frac{1}{3} \int \sqrt{z+1} \, dz - \frac{1}{3} \int (z+1)^{-1/2} \, dz\)
\(= \frac{1}{3} \cdot \frac{(z+1)^{3/2}}{3/2} - \frac{1}{3} \cdot \frac{(z+1)^{1/2}}{1/2} + C\)
\(= \frac{2}{9} (z+1)^{3/2} - \frac{2}{3} (z+1)^{1/2} + C\)
\(= \frac{2}{9} (1 + x^{3})^{3/2} - \frac{2}{3} (1 + x^{3})^{1/2} + C \quad (Ans)\)
Solution:
\(I = \int \frac{\sqrt{\tan x}}{\sin 2x} \, dx\)
\(= \int \frac{\sqrt{\tan x}}{2 \sin x \cos x} \, dx\)
\(= \frac{1}{2} \int \frac{\sqrt{\frac{\sin x}{\cos x}}}{\sin x \cos x} \, dx\)
\(= \frac{1}{2} \int \frac{\sqrt{\sin x}}{\sqrt{\cos x} \sin x \cos x} \, dx\)
\(= \frac{1}{2} \int \frac{dx}{\sqrt{\tan x} \cos^{2} x}\)
\(= \frac{1}{2} \int \frac{\sec^{2} x \, dx}{\sqrt{\tan x}}\)
Let, \(\; \tan x = z\)
\(\sec^{2} x \, dx = dz\)
\(I = \frac{1}{2} \int \frac{dz}{\sqrt{z}}\)
\(= \frac{1}{2} \int z^{-1/2} dz\)
\(= \frac{1}{2} \cdot \frac{z^{1/2}}{1/2} + C\)
\(= \sqrt{z} + C\)
\(= \sqrt{\tan x} + C \quad (Answer)\)
Solution:
\(I = \int \frac{dx}{1 - e^{x}}\)
\(= \int \frac{e^{x} + (1 - e^{x})}{1 - e^{x}} \, dx\)
\(= \int \frac{e^{x} dx}{1 - e^{x}} + \int dx\)
\(= \frac{\ln |1 - e^{x}|}{-1} + x + C\)
\(= - \ln |1 - e^{x}| + x + C \quad (Ans)\)
Solution:
\(I = \int \frac{x e^{x}}{(x+1)^{2}} \, dx\)
\(= \int \frac{(x e^{x} + e^{x}) - e^{x}}{(x+1)^{2}} \, dx\)
\(= \int \frac{e^{x}(x+1)}{(x+1)^{2}} \, dx - \int \frac{e^{x}}{(x+1)^{2}} \, dx\)
\(= \int \frac{e^{x}}{x+1} \, dx - \int \frac{e^{x}}{(x+1)^{2}} \, dx\)
\(= \frac{1}{x+1} \int e^{x} dx - \int \frac{d}{dx}\left(\frac{1}{x+1}\right) e^{x} dx - \int \frac{e^{x}}{(x+1)^{2}} dx\)
\(= \frac{e^{x}}{x+1} + \int \frac{e^{x}}{(x+1)^{2}} dx - \int \frac{e^{x}}{(x+1)^{2}} dx + C\)
\(= \frac{e^{x}}{x+1} + C \quad (Ans)\)
Solution:
\(I = \int x e^{x} \, dx\)
\(= x \int e^{x} dx - \int \left[ \frac{d}{dx}(x) \, e^{x} \right] dx + C\)
\(= x e^{x} - \int e^{x} dx\)
\(= x e^{x} - e^{x} + C \quad (Ans)\)
Solution:
\(I = \int \log x \, dx\)
\(= \int \log x \cdot (1) \, dx\)
\(= \log x \int dx - \int \left[ \frac{d}{dx}(\log x) \int dx \right] dx + C\)
\(= x \log x - \int \frac{1}{x} \cdot x \, dx + C\)
\(= x \log x - x + C \quad (Ans)\)
Solution:
\(I = \int \frac{dx}{x^{2} + 6x + 10}\)
\(x^{2} + 6x + 10\)
\(= (x^{2} + 2 \cdot x \cdot 3 + 3^{2}) + 10 - 9\)
\(= (x+3)^{2} + (1)^{2}\)
Now, \(\; I = \int \frac{dx}{(x+3)^{2} + 1^{2}}\)
\(= \frac{1}{1} \tan^{-1} \left( \frac{x+3}{1} \right) + C\)
\(= \tan^{-1}(x+3) + C \quad (Ans)\)
Solution:
\(\displaystyle \int_{-1}^{1} |x| \, dx\)
\(|x| = \begin{cases} x & 0 \le x \le 1 \\ -x & -1 \le x < 0 \end{cases}\)
Hence, \(\displaystyle \int_{-1}^{1} |x| \, dx\)
\(= \int_{-1}^{0} -x \, dx + \int_{0}^{1} x \, dx\)
\(= -\frac{1}{2} [x^{2}]_{-1}^{0} + \frac{1}{2} [x^{2}]_{0}^{1}\)
\(= -\frac{1}{2} (0 - (-1)^{2}) + \frac{1}{2} (1^{2} - 0^{2})\)
\(= \left(-\frac{1}{2}\right)(-1) + \frac{1}{2} \times 1\)
\(= \frac{1}{2} + \frac{1}{2} = 1 \quad (Ans)\)
Solution:
\(I = \int_{0}^{\log 2} \frac{e^{x}}{1 + e^{x}} \, dx\)
\(= \left[ \ln |1 + e^{x}| \right]_{0}^{\log 2}\)
\(= \ln |1 + e^{\log 2}| - \ln |1 + e^{0}|\)
\(= \ln |1 + 2| - \ln |1 + 1|\)
\(= \log 3 - \log 2\)
\(= \log \left( \frac{3}{2} \right) \quad (Ans)\)
Solution:
\(I = \int_{0}^{\pi/2} (a\cos^{2}x + b\sin^{2}x)\, dx\)
As \(\int_{0}^{a} f(x)\, dx = \int_{0}^{a} f(a-x)\, dx\)
So,
\(\Rightarrow I = \int_{0}^{\pi/2} \{ a\cos^{2}(\pi/2 - x) + b\sin^{2}(\pi/2 - x) \} dx\)
\(= \int_{0}^{\pi/2} (a\sin^{2}x + b\cos^{2}x)\, dx\)
Now, \(I + I = \int_{0}^{\pi/2} (a\cos^{2}x + b\sin^{2}x)\, dx + \int_{0}^{\pi/2} (a\sin^{2}x + b\cos^{2}x)\, dx\)
\(\Rightarrow 2I = \int_{0}^{\pi/2} [a(\cos^{2}x + \sin^{2}x) + b(\sin^{2}x + \cos^{2}x)] dx\)
\(\Rightarrow I = \frac{1}{2} \int_{0}^{\pi/2} (a + b)\, dx\)
\(= \frac{a+b}{2} [x]_{0}^{\pi/2}\)
\(= \frac{(a+b)\pi}{4} \quad (Ans)\)
Solution:
\(I = \int_{-1}^{1} (x + |x|)\, dx\)
\(x + |x| = \begin{cases} x + x = 2x & 0 \le x \le 1 \\ x - x = 0 & -1 \le x < 0 \end{cases}\)
So, \(I = \int_{-1}^{0} 0\, dx + \int_{0}^{1} 2x\, dx\)
\(= 0 + 2 \left[ \frac{x^{2}}{2} \right]_{0}^{1}\)
\(= 1^{2} - 0^{2}\)
\(= 1 \quad (Ans)\)
Solution:
\(I = \int_{-1}^{1} x|x| \, dx\)
\(x|x| = \begin{cases} x(+x) = x^{2} & 0 \le x \le 1 \\ x(-x) = -x^{2} & -1 \le x < 0 \end{cases}\)
So, \(I = \int_{-1}^{0} -x^{2} dx + \int_{0}^{1} x^{2} dx\)
\(= -\frac{1}{3} [x^{3}]_{-1}^{0} + \frac{1}{3} [x^{3}]_{0}^{1}\)
\(= -\frac{1}{3} (0^{3} - (-1)^{3}) + \frac{1}{3} (1^{3} - 0^{3})\)
\(= -\frac{1}{3} (+1) + \frac{1}{3} \times 1\)
\(= -\frac{1}{3} + \frac{1}{3} = 0 \quad (Ans)\)
Solution:
\(I = \int_{0}^{3} |x^{2}-1| \, dx\)
For zeros of \((x^{2}-1)\), we have, \(x^{2}-1=0 \Rightarrow x=\pm 1\)
\(|x^{2}-1| = \begin{cases} + (x^{2}-1) & 1 \le x \le 3 \\ - (x^{2}-1) & 0 \le x < 1 \end{cases}\)
So, \(I = \int_{0}^{1} -(x^{2}-1)\, dx + \int_{1}^{3} (x^{2}-1)\, dx\)
\(= -\frac{1}{3} [x^{3}]_{0}^{1} + [x]_{0}^{1} + \frac{1}{3} [x^{3}]_{1}^{3} - [x]_{1}^{3}\)
\(= -\frac{1}{3}(1) + (1-0) + \frac{1}{3}(3^{3}-1^{3}) - (3-1)\)
\(= -\frac{1}{3} + 1 + \frac{1}{3}(27-1) - 2\)
\(= -\frac{1}{3} + \frac{26}{3} - 1\)
\(= \frac{22}{3} \quad (Ans)\)
Solution:
\(I = \int_{1/e}^{e} |\log x| \, dx\)
\(|\log x| = \begin{cases} + \log x & 1 \le x \le e \\ - \log x & \frac{1}{e} \le x < 1 \end{cases}\)
So, \(I = \int_{1/e}^{1} -\log x \, dx + \int_{1}^{e} \log x \, dx\)
As, \(\int \log x \, dx = x \log x - x + C\)
\(I = -[x\log x - x]_{1/e}^{1} + [x\log x - x]_{1}^{e}\)
\(= -[(1\cdot0 - 1) - (\frac{1}{e}(-1) - \frac{1}{e})] + [(e\cdot1 - e) - (1\cdot0 - 1)]\)
\(= -[-1 - (-\frac{1}{e} - \frac{1}{e})] + [(e-e) - (-1)]\)
\(= -[-1 + \frac{2}{e}] + [0 + 1]\)
\(= 1 - \frac{2}{e} + 1\)
\(= 2 - \frac{2}{e} \quad (Ans)\)
Solution:
\(I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx\)
As, \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\)
\(\Rightarrow I = \int_{0}^{\pi/2} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} \, dx\)
\(= \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx\)
Now, \(I + I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx + \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx\)
\(\Rightarrow 2I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx\)
\(\Rightarrow 2I = \int_{0}^{\pi/2} 1 \, dx\)
\(\Rightarrow I = \frac{1}{2} \int_{0}^{\pi/2} dx\)
\(= \frac{1}{2} [x]_{0}^{\pi/2}\)
\(= \frac{\pi}{4} \quad (Ans)\)
Solution:
\(I = \int_{1}^{3} |x - 2| \, dx\)
\(|x - 2| = \begin{cases} -(x - 2) & 1 \le x < 2 \\ +(x - 2) & 2 \le x < 3 \end{cases}\)
So, \(I = \int_{1}^{2} -(x - 2)\, dx + \int_{2}^{3} (x - 2)\, dx\)
\(= -\frac{1}{2} [x^{2}]_{1}^{2} + 2[x]_{1}^{2} + \frac{1}{2} [x^{2}]_{2}^{3} - 2[x]_{2}^{3}\)
\(= -\frac{1}{2} (4 - 1) + 2(2 - 1) + \frac{1}{2} (9 - 4) - 2(3 - 2)\)
\(= -\frac{3}{2} + 2 + \frac{5}{2} - 2\)
\(= 1 \quad (Ans)\)
Solution:
\(I = \int_{-3}^{3} \frac{x e^{x^{2}}}{1 + x^{2}} \, dx\)
Let, \(f(x) = \frac{x e^{x^{2}}}{1 + x^{2}}\)
\(\Rightarrow f(-x) = \frac{-x e^{(-x)^{2}}}{1 + (-x)^{2}}\)
\(= \frac{-x e^{x^{2}}}{1 + x^{2}}\)
\(= -f(x)\)
So, \(f(x)\) is an odd function.
\(\therefore I = 0 \quad [\text{As, } \int_{-a}^{a} f(x)\,dx = 0 \text{ if } f(x) \text{ is odd}]\)
Solution:
\(I = \int_{0}^{\pi/2} \log(\cot x)\, dx\)
As, \(\int_{0}^{a} f(x)\, dx = \int_{0}^{a} f(a-x)\, dx\)
So, \(I = \int_{0}^{\pi/2} \log[\cot(\pi/2 - x)]\, dx\)
\(= \int_{0}^{\pi/2} \log(\tan x)\, dx\)
\(= \int_{0}^{\pi/2} \log\left(\frac{1}{\cot x}\right)\, dx\)
\(= - \int_{0}^{\pi/2} \log(\cot x)\, dx\)
\(= -I\)
\(\Rightarrow 2I = 0 \Rightarrow I = 0 \quad (Ans)\)
Solution:
\(I = \int_{0}^{\pi} x f(\sin x)\, dx\)
As, \(\int_{0}^{a} f(x)\, dx = \int_{0}^{a} f(a-x)\, dx\)
So, \(I = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x))\, dx\)
\(= \int_{0}^{\pi} (\pi - x) f(\sin x)\, dx\)
Now, \(I + I = \int_{0}^{\pi} x f(\sin x)\, dx + \int_{0}^{\pi} (\pi - x) f(\sin x)\, dx\)
\(\Rightarrow 2I = \int_{0}^{\pi} \pi f(\sin x)\, dx\)
\(= \pi \int_{0}^{\pi} f(\sin x)\, dx\)
As, \(\sin(\pi - x) = \sin x\)
\(\therefore \int_{0}^{\pi} f(\sin x)\, dx = 2 \int_{0}^{\pi/2} f(\sin x)\, dx\)
Hence, \(2I = 2\pi \int_{0}^{\pi/2} f(\sin x)\, dx\)
\(\Rightarrow I = \pi \int_{0}^{\pi/2} f(\sin x)\, dx \quad (Ans)\)
Solution:
As we have \(\displaystyle \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a+b-x)\,dx\)
So, \(I = \int_{a}^{b} \frac{f(x)\,dx}{f(x) + f(a+b-x)}\)
\(= \int_{a}^{b} \frac{f(a+b-x)\,dx}{f(a+b-x) + f(a+b-(a+b-x))}\)
\(= \int_{a}^{b} \frac{f(a+b-x)\,dx}{f(a+b-x) + f(x)}\)
Now, \(I + I = \int_{a}^{b} \frac{f(x) + f(a+b-x)}{f(x) + f(a+b-x)}\, dx\)
\(\Rightarrow 2I = \int_{a}^{b} dx\)
\(\Rightarrow I = \frac{1}{2} [x]_{a}^{b}\)
\(= \frac{b-a}{2}\)
Hence, \(\displaystyle \int_{a}^{b} \frac{f(x)\,dx}{f(x) + f(a+b-x)} = \frac{b-a}{2} \quad (Answer)\)
Solution:
As, \(\displaystyle \int_{0}^{a} f(x)\, dx = \int_{0}^{a} f(a-x)\, dx\)
So, \(I = \int_{0}^{10} \frac{x^{3}}{x^{3} + (10-x)^{3}} \, dx\)
\(= \int_{0}^{10} \frac{(10-x)^{3}}{(10-x)^{3} + x^{3}} \, dx\)
Now, \(I + I = \int_{0}^{10} \frac{x^{3} + (10-x)^{3}}{x^{3} + (10-x)^{3}} \, dx\)
\(\Rightarrow 2I = \int_{0}^{10} dx\)
\(\Rightarrow I = \frac{1}{2} [x]_{0}^{10}\)
\(= 5\)
Hence, \(\displaystyle \int_{0}^{10} \frac{x^{3}}{x^{3} + (10-x)^{3}} \, dx = 5 \quad (Ans)\)
Solution:
\(I = \int \frac{\sin^{6}x}{\cos^{8}x} \, dx\)
\(= \int \tan^{6}x \sec^{2}x \, dx\)
Let, \(\tan x = z\)
Differentiating w.r.t. x, we get
\(\sec^{2}x \, dx = dz\)
So, \(I = \int z^{6} dz\)
\(= \frac{z^{7}}{7} + C\)
\(= \frac{\tan^{7}x}{7} + C \quad (Answer)\)
Solution:
\(I = \int \frac{dx}{x\sqrt{x+1}}\)
Let, \(x+1 = z^{2} \Rightarrow x = z^{2} - 1\)
Differentiating w.r.t. x, we get
\(dx = 2z\, dz\)
So, \(I = \int \frac{2z\, dz}{(z^{2}-1)z}\)
\(= 2 \int \frac{dz}{z^{2}-1}\)
\(= 2 \times \frac{1}{2 \times 1} \log \left| \frac{z-1}{z+1} \right| + C\)
\(= \log \left| \frac{z-1}{z+1} \right| + C\)
\(= \log \left| \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right| + C \quad (Answer)\)
Solution:
\(I = \int \frac{\cos x + x\sin x}{x(x+\cos x)} \, dx\)
\(= \int \frac{(x+\cos x) + (x\sin x - x)}{x(x+\cos x)} \, dx\)
\(= \int \frac{dx}{x} + \int \frac{x(\sin x - 1)}{x(x+\cos x)} \, dx\)
\(= \int \frac{dx}{x} - \int \frac{1-\sin x}{x+\cos x} \, dx\)
\(= \ln|x| - \ln|x+\cos x| + C\)
\(= -\ln\left|\frac{x}{x+\cos x}\right| + C \quad (Ans)\)
📗 3 Mark Integrals
30+ problemsSolution:
Let \(x = a\tan^{2}\theta\)
Differentiating w.r.t. \(x\), we get
\(dx = 2a\tan\theta \sec^{2}\theta \, d\theta\)
Now, \(I = \int \sin^{-1}\!\sqrt{\frac{a\tan^{2}\theta}{a+a\tan^{2}\theta}} \cdot 2a\tan\theta \sec^{2}\theta \, d\theta\)
\(= \int \sin^{-1}\!\sqrt{\frac{\tan^{2}\theta}{\sec^{2}\theta}} \cdot 2a\tan\theta \sec^{2}\theta \, d\theta\)
\(= \int \sin^{-1}(\sin\theta) \cdot 2a\tan\theta \sec^{2}\theta \, d\theta\)
\(= 2a \int \theta \tan\theta \sec^{2}\theta \, d\theta\)
\(= 2a \left[ \theta \int \tan\theta \sec^{2}\theta \, d\theta - \int \theta \cdot \tan\theta \sec^{2}\theta \, d\theta \right] + C\)
\(= 2a \left[ \theta \cdot \frac{\tan^{2}\theta}{2} - \int \frac{\tan^{2}\theta}{2} \, d\theta \right] + C\)
\(= a\theta \tan^{2}\theta - a \int \tan^{2}\theta \, d\theta + C\)
\(= a\theta \tan^{2}\theta - a \int (\sec^{2}\theta - 1)\, d\theta + C\)
\(= a\theta \tan^{2}\theta - a\tan\theta + a\theta + C\)
\(= x \tan^{-1}\!\sqrt{\frac{x}{a}} - a\sqrt{\frac{x}{a}} + a\tan^{-1}\!\sqrt{\frac{x}{a}} + C\)
\(= (x+a)\tan^{-1}\!\sqrt{\frac{x}{a}} - \sqrt{ax} + C \quad (Ans)\)
Solution:
Let, \(x^{2} = a^{2}\cos 2\theta\)
Differentiating w.r.t. \(x\), we get
\(2x\,dx = a^{2}(-\sin 2\theta)\, d\theta\)
\(x\,dx = -a^{2}\sin 2\theta\, d\theta\)
So, \(I = \int \sqrt{\frac{a^{2}-a^{2}\cos 2\theta}{a^{2}+a^{2}\cos 2\theta}} \, (x\,dx)\)
\(= \int \sqrt{\frac{1-\cos 2\theta}{1+\cos 2\theta}} \, (-a^{2}\sin 2\theta)\, d\theta\)
\(= \int \tan\theta \, (-a^{2}\cdot 2\sin\theta\cos\theta)\, d\theta\)
\(= -\int 2a^{2}\sin^{2}\theta\, d\theta\)
\(= -a^{2}\int (1-\cos 2\theta)\, d\theta\)
\(= -a^{2}\theta + \frac{a^{2}\sin 2\theta}{2} + C\)
\(= -a^{2}\cos^{-1}\!\left(\frac{x^{2}}{a^{2}}\right) + \frac{a^{2}}{2}\sqrt{1-\frac{x^{4}}{a^{4}}} + C \quad (Ans)\)
\(\left[\because \sin 2\theta = \sqrt{1-\cos^{2}2\theta} = \sqrt{1-\frac{x^{4}}{a^{4}}}\right]\)
Solution:
\(I = \int e^{x}\frac{x^{2}+1}{(x+1)^{2}} \, dx\)
\(= \int e^{x}\frac{(x^{2}-1)+2}{(x+1)^{2}} \, dx\)
\(= \int e^{x}\frac{(x+1)(x-1)+2}{(x+1)^{2}} \, dx\)
\(= \int e^{x}\frac{x-1}{x+1} \, dx + \int \frac{2e^{x}}{(x+1)^{2}} \, dx\)
\(= \left(\frac{x-1}{x+1}\right)e^{x} - \int \frac{d}{dx}\!\left(\frac{x-1}{x+1}\right)e^{x} \, dx + \int \frac{2e^{x}}{(x+1)^{2}} \, dx + C\)
\(= e^{x}\frac{x-1}{x+1} - \int \frac{2e^{x}}{(x+1)^{2}} \, dx + \int \frac{2e^{x}}{(x+1)^{2}} \, dx + C\)
\(= e^{x}\frac{x-1}{x+1} + C \quad (Ans)\)
Solution:
\(I = \int \frac{dx}{\cos x - \sqrt{3}\sin x}\)
\(= \int \frac{\frac{1}{2}\,dx}{\frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x}\)
\(= \frac{1}{2}\int \frac{dx}{\cos x\cos\frac{\pi}{3} - \sin x\sin\frac{\pi}{3}}\)
\(= \frac{1}{2}\int \frac{dx}{\cos\left(x+\frac{\pi}{3}\right)}\)
\(= \frac{1}{2}\int \sec\left(x+\frac{\pi}{3}\right)\,dx\)
\(= \frac{1}{2}\log\left|\sec\left(x+\frac{\pi}{3}\right)+\tan\left(x+\frac{\pi}{3}\right)\right|+C \quad (Ans)\)
Solution:
\(I = \int \frac{dx}{a^{2}\cos^{2}x + b^{2}\sin^{2}x}\)
\(= \int \frac{\sec^{2}x\,dx}{a^{2} + b^{2}\tan^{2}x}\)
Let, \(\tan x = z\)
\(\sec^{2}x\,dx = dz\)
So, \(I = \int \frac{dz}{a^{2} + b^{2}z^{2}}\)
\(= \frac{1}{a^{2}}\int \frac{dz}{1 + \left(\frac{b}{a}z\right)^{2}}\)
\(= \frac{1}{a^{2}} \cdot \frac{a}{b}\tan^{-1}\!\left(\frac{bz}{a}\right) + C\)
\(= \frac{1}{ab}\tan^{-1}\!\left(\frac{b\tan x}{a}\right) + C \quad (Ans)\)
Solution:
As, \(\displaystyle \cos x = \frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\)
So, \(I = \int \frac{dx}{5 + 4\left(\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)}\)
\(= \int \frac{(1+\tan^{2}\frac{x}{2})\,dx}{5(1+\tan^{2}\frac{x}{2}) + 4(1-\tan^{2}\frac{x}{2})}\)
\(= \int \frac{\sec^{2}\frac{x}{2}\,dx}{9 + \tan^{2}\frac{x}{2}}\)
Let, \(\tan\frac{x}{2} = z\)
\(\frac{1}{2}\sec^{2}\frac{x}{2}\,dx = dz \Rightarrow \sec^{2}\frac{x}{2}\,dx = 2dz\)
So, \(I = \int \frac{2\,dz}{9 + z^{2}}\)
\(= 2 \cdot \frac{1}{3}\tan^{-1}\!\left(\frac{z}{3}\right) + C\)
\(= \frac{2}{3}\tan^{-1}\!\left(\frac{\tan\frac{x}{2}}{3}\right) + C \quad (Ans)\)
Solution:
As, \(\displaystyle \sin x = \frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\)
So, \(I = \int \frac{dx}{5 - 4\left(\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)}\)
\(= \int \frac{(1+\tan^{2}\frac{x}{2})\,dx}{5(1+\tan^{2}\frac{x}{2}) - 8\tan\frac{x}{2}}\)
\(= \int \frac{\sec^{2}\frac{x}{2}\,dx}{5\tan^{2}\frac{x}{2} - 8\tan\frac{x}{2} + 5}\)
Let, \(\tan\frac{x}{2} = z\)
\(\frac{1}{2}\sec^{2}\frac{x}{2}\,dx = dz \Rightarrow \sec^{2}\frac{x}{2}\,dx = 2dz\)
Now, \(I = \int \frac{2\,dz}{5z^{2} - 8z + 5}\)
\(= \frac{2}{5}\int \frac{dz}{z^{2} - \frac{8}{5}z + 1}\)
\(= \frac{2}{5}\int \frac{dz}{\left(z-\frac{4}{5}\right)^{2} + \left(\frac{3}{5}\right)^{2}}\)
\(= \frac{2}{5}\cdot \frac{5}{3}\tan^{-1}\!\left(\frac{z-\frac{4}{5}}{\frac{3}{5}}\right) + C\)
\(= \frac{2}{3}\tan^{-1}\!\left(\frac{5z-4}{3}\right) + C\)
\(= \frac{2}{3}\tan^{-1}\!\left(\frac{5\tan\frac{x}{2}-4}{3}\right) + C \quad (Ans)\)
Solution:
\(I = \int \frac{2\cos x + 3\sin x}{3\cos x + 2\sin x}\, dx\)
Let, \(2\cos x + 3\sin x = m(3\cos x + 2\sin x) + n\frac{d}{dx}(3\cos x + 2\sin x)\)
\(= 3m\cos x + 2m\sin x + n(-3\sin x + 2\cos x)\)
\(= (3m + 2n)\cos x + (2m - 3n)\sin x\)
So, \(3m + 2n = 2 \quad \text{and} \quad 2m - 3n = 3\)
\(\Rightarrow m = \frac{2-2n}{3}\)
\(\Rightarrow 2\left(\frac{2-2n}{3}\right) - 3n = 3\)
\(\Rightarrow 4 - 4n - 9n = 9\)
\(\Rightarrow 13n = -5 \Rightarrow n = -\frac{5}{13}\)
\(\Rightarrow m = \frac{2-2\left(-\frac{5}{13}\right)}{3} = \frac{12}{13}\)
Now, \(I = \int \frac{m(3\cos x + 2\sin x) + n\frac{d}{dx}(3\cos x + 2\sin x)}{3\cos x + 2\sin x}\, dx\)
\(= m\int dx + n\int \frac{d(3\cos x + 2\sin x)}{3\cos x + 2\sin x}\)
\(= mx + n\ln|3\cos x + 2\sin x| + C\)
\(= \frac{12}{13}x - \frac{5}{13}\ln|3\cos x + 2\sin x| + C \quad (Ans)\)
Solution:
\(I = \int \frac{\sin 2x}{\sin^{2}x + \cos^{4}x}\, dx\)
Let, \(\sin^{2}x = z\)
\(\Rightarrow 2\sin x\cos x\, dx = dz\)
\(\Rightarrow \sin 2x\, dx = dz\)
So, \(I = \int \frac{dz}{z^{2} - z + 1}\)
\(= \int \frac{dz}{\left(z-\frac{1}{2}\right)^{2} + \left(\frac{\sqrt{3}}{2}\right)^{2}}\)
\(= \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\!\left(\frac{z-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C\)
\(= \frac{2}{\sqrt{3}} \tan^{-1}\!\left(\frac{2z-1}{\sqrt{3}}\right) + C\)
\(= \frac{2}{\sqrt{3}} \tan^{-1}\!\left(\frac{2\sin^{2}x-1}{\sqrt{3}}\right) + C \quad (Ans)\)
Solution:
\(I = \int x(\tan^{-1}x)^{2}\, dx\)
\(= (\tan^{-1}x)^{2}\int x\,dx - \int \frac{d}{dx}(\tan^{-1}x)^{2}\left(\int x\,dx\right)dx + C\)
\(= \frac{x^{2}}{2}(\tan^{-1}x)^{2} - \int 2\tan^{-1}x\frac{1}{1+x^{2}}\frac{x^{2}}{2}\,dx + C\)
\(= \frac{x^{2}}{2}(\tan^{-1}x)^{2} - \int \frac{x^{2}\tan^{-1}x}{1+x^{2}}\,dx + C\)
Let, \(x=\tan\theta\)
\(dx=\sec^{2}\theta\,d\theta\)
\(\frac{x^{2}\tan^{-1}x}{1+x^{2}}=\frac{\tan^{2}\theta\cdot\theta}{\sec^{2}\theta} =\theta\sin^{2}\theta\)
So, \(I=\frac{x^{2}}{2}(\tan^{-1}x)^{2}-\int \theta\sin^{2}\theta\,d\theta + C\)
\(=\frac{x^{2}}{2}(\tan^{-1}x)^{2} -\int \theta(\sec^{2}\theta-1)\,d\theta + C\)
\(=\frac{x^{2}}{2}(\tan^{-1}x)^{2} -\int \theta\sec^{2}\theta\,d\theta +\int \theta\,d\theta + C\)
\(=\frac{x^{2}}{2}(\tan^{-1}x)^{2} +\frac{1}{2}(\tan^{-1}x)^{2} -x\tan^{-1}x +\ln|\sec\theta| + C\)
\(=\frac{x^{2}}{2}(\tan^{-1}x)^{2} +\frac{(\tan^{-1}x)^{2}}{2} -x\tan^{-1}x +\ln\sqrt{1+x^{2}} + C\)
Solution:
\(I = \int (\log x)^{2}\, dx\)
\(= \int (\log x)^{2}\cdot(1)\, dx\)
\(= (\log x)^{2}\int dx - \int \frac{d}{dx}(\log x)^{2}\left(\int dx\right)dx + C\)
\(= x(\log x)^{2} - \int 2\log x\cdot\frac{1}{x}\cdot x\,dx + C\)
\(= x(\log x)^{2} - 2\int \log x\,dx + C\)
\(= x(\log x)^{2} - 2(x\log x - x) + C\)
\(= x(\log x)^{2} - 2x\log x + 2x + C \quad (Ans)\)
Solution:
\(I = \int \frac{\log x}{(1+\log x)^{2}}\, dx\)
\(= \int \frac{(1+\log x)-1}{(1+\log x)^{2}}\, dx\)
\(= \int \left(\frac{1}{1+\log x}-\frac{1}{(1+\log x)^{2}}\right) dx\)
Now, let \(1+\log x=z\Rightarrow \log x=z-1\Rightarrow x=e^{z-1}\)
Differentiating w.r.t. \(x\), we get
\(\frac{dx}{x}=dz\Rightarrow dx=x\,dz\)
\(dx=e^{z-1}dz\)
So, \(I=\int\left(\frac{1}{z}-\frac{1}{z^{2}}\right)e^{z-1}dz\)
\(=\frac{1}{e}\int e^{z}\left(\frac{1}{z}-\frac{1}{z^{2}}\right)dz\)
We have, \(\int e^{z}\left(f(z)+f'(z)\right)dz=e^{z}f(z)+C\)
For our case, \(f(z)=\frac{1}{z}\), \(f'(z)=-\frac{1}{z^{2}}\)
So, \(I=\frac{1}{e}\cdot e^{z}\left(\frac{1}{z}\right)+C\)
\(=\frac{e^{z-1}}{z}+C\)
\(=\frac{x}{1+\log x}+C\)
Hence, \(\displaystyle \int \frac{\log x}{(1+\log x)^{2}}\, dx=\frac{x}{1+\log x}+C\quad (Answer)\)
Solution:
Let \(x=a\cos 2\theta\)
Differentiating w.r.t. \(x\), we get
\(dx=a(-2\sin 2\theta)\, d\theta\)
and
\(\sqrt{\frac{a+x}{a-x}} =\sqrt{\frac{a+a\cos 2\theta}{a-a\cos 2\theta}} =\sqrt{\frac{1+\cos 2\theta}{1-\cos 2\theta}} =\cot\theta\)
\(\left[\because \tan^{2}\theta=\frac{1-\cos 2\theta}{1+\cos 2\theta}\right]\)
So, \(I=\int \cot\theta\cdot(-2a\sin 2\theta)\, d\theta\)
\(=-\int \cot\theta\cdot 2a\cdot 2\sin\theta\cos\theta\, d\theta\)
\(=-2a\int 2\cos^{2}\theta\, d\theta\)
\(=-2a\int(1+\cos 2\theta)\, d\theta\)
\(=-2a\theta+a\sin 2\theta+C\)
\(=-a\cos^{-1}\!\left(\frac{x}{a}\right) +a\sqrt{1-\frac{x^{2}}{a^{2}}}+C\quad (Ans)\)
Solution:
\(I=\int \frac{dx}{(1+x)\sqrt{1-x^{2}}}\)
Let \(x=\sin\theta\)
\(dx=\cos\theta\,d\theta\)
So, \(I=\int \frac{\cos\theta\,d\theta}{(1+\sin\theta)\cos\theta}\)
\(=\int \frac{d\theta}{1+\sin\theta}\)
\(=\int \frac{1-\sin\theta}{\cos^{2}\theta}\,d\theta\)
\(=\int \sec^{2}\theta\,d\theta-\int \sec\theta\tan\theta\,d\theta\)
\(=\tan\theta-\sec\theta+C\)
\(=\frac{x}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}+C\quad (Ans)\)
Solution:
\(I=\int \frac{\sec^{2}x\,dx}{\sqrt{5\tan^{2}x-12\tan x+4}}\)
Let \(\tan x=z\)
\(\sec^{2}x\,dx=dz\)
So, \(I=\int \frac{dz}{\sqrt{5z^{2}-12z+4}}\)
\(=\frac{1}{\sqrt{5}}\int \frac{dz}{\sqrt{z^{2}-\frac{12}{5}z+\frac{4}{5}}}\)
\(=\frac{1}{\sqrt{5}}\int \frac{dz}{\sqrt{\left(z-\frac{6}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}}}\)
\(=\frac{1}{\sqrt{5}} \ln\left|z-\frac{6}{5} +\sqrt{\left(z-\frac{6}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}}\right| +C\)
\(=\frac{1}{\sqrt{5}} \ln\left|\tan x-\frac{6}{5} +\sqrt{\left(\tan x-\frac{6}{5}\right)^{2}-\left(\frac{4}{5}\right)^{2}}\right| +C \quad (Ans)\)
Solution:
\(I=\int \frac{dx}{\sqrt{2ax-x^{2}}}\)
\(= \int \frac{dx}{\sqrt{-\left(x^{2}-2ax+a^{2}-a^{2}\right)}}\)
\(= \int \frac{dx}{\sqrt{a^{2}-(x-a)^{2}}}\)
\(=\sin^{-1}\!\left(\frac{x-a}{a}\right)+C\quad (Ans)\)
Solution:
\(I=\int \frac{x+2}{x^{2}-3x+1}\, dx\)
\(=\int \frac{\frac{1}{2}(2x-3)+\frac{7}{2}}{x^{2}-3x+1}\, dx\)
\(=\frac{1}{2}\int \frac{2x-3}{x^{2}-3x+1}\, dx +\frac{7}{2}\int \frac{dx}{x^{2}-3x+1}\)
\(=\frac{1}{2}\ln|x^{2}-3x+1| +\frac{7}{2}\int \frac{dx}{(x-\frac{3}{2})^{2}-\frac{5}{4}}+C\)
\(=\frac{1}{2}\ln|x^{2}-3x+1| +\frac{7}{2}\int \frac{dx}{\left(\frac{\sqrt5}{2}\right)^{2}-(x-\frac{3}{2})^{2}}+C\)
\(=\frac{1}{2}\ln|x^{2}-3x+1| +\frac{7}{2}\cdot\frac{1}{2\cdot\frac{\sqrt5}{2}} \ln\left|\frac{x-\frac{3}{2}+\frac{\sqrt5}{2}} {-\;x+\frac{3}{2}+\frac{\sqrt5}{2}}\right|+C\)
\(=\frac{1}{2}\ln|x^{2}-3x+1| +\frac{7}{2\sqrt5} \ln\left|\frac{2x-3+\sqrt5}{-2x+3+\sqrt5}\right| +C \quad (Ans)\)
Solution:
\(I=\int \frac{dx}{\sin(x-a)\sin(x-b)}\)
\(=\frac{1}{\sin(a-b)}\int \frac{\sin(x-b)-\sin(x-a)}{\sin(x-a)\sin(x-b)}\,dx\)
\(=\frac{1}{\sin(a-b)} \int\left(\frac{\cos(x-a)}{\sin(x-a)} -\frac{\cos(x-b)}{\sin(x-b)}\right)dx\)
\(=\frac{1}{\sin(a-b)} \left(\int\frac{\cos(x-a)}{\sin(x-a)}dx -\int\frac{\cos(x-b)}{\sin(x-b)}dx\right)\)
\(=\frac{1}{\sin(a-b)} \left(\ln|\sin(x-a)| -\ln|\sin(x-b)|\right)+C\)
\(=\frac{1}{\sin(a-b)} \ln\left|\frac{\sin(x-a)}{\sin(x-b)}\right| +C \quad (Ans)\)
Solution:
Let \(a=r\cos\alpha,\; b=r\sin\alpha\)
\(\therefore a^{2}+b^{2}=r^{2}\)
and \(\tan\alpha=\frac{b}{a}\)
Now, \(I=\int \frac{dx}{r(\sin x\cos\alpha+\cos x\sin\alpha)}\)
\(=\frac{1}{r}\int \frac{dx}{\sin(x+\alpha)}\)
\(=\frac{1}{r}\int \text{cosec}(x+\alpha)\,dx\)
\(=\frac{1}{r}\ln\left|\text{cosec}(x+\alpha)-\cot(x+\alpha)\right|+C\)
\(=\frac{1}{\sqrt{a^{2}+b^{2}}} \ln\left|\text{cosec}\!\left(x+\tan^{-1}\frac{b}{a}\right) -\cot\!\left(x+\tan^{-1}\frac{b}{a}\right)\right| +C \quad (Ans)\)
Solution:
Let \(\log x=z \Rightarrow e^{z}=x\)
Differentiating w.r.t. \(x\), we get
\(dx=e^{z}dz\)
Now, \(I=\int \left(\log z+\frac{1}{z}\right)e^{z}dz\)
\(=\int e^{z}\log z\,dz+\int \frac{e^{z}}{z}dz\)
\(=e^{z}\log z-\int \frac{e^{z}}{z}dz+\int \frac{e^{z}}{z}dz+C\)
\(=e^{z}\log z+C\)
\(=x\log(\log x)+C \quad (Ans)\)
Solution:
Let \(x^{\frac{1}{4}}=z\)
\(\Rightarrow x=z^{4}\)
\(\Rightarrow \sqrt{x}=z^{2}\)
Differentiating w.r.t. \(x\), we get
\(dx=4z^{3}dz\)
So, \(I=\int \frac{z^{2}\cdot4z^{3}dz}{1+z^{3}}\)
\(=4\int \frac{z^{5}}{1+z^{3}}dz\)
\(\frac{z^{5}}{1+z^{3}} =z^{2}-\frac{z^{2}}{1+z^{3}}\)
So, \(I=4\int z^{2}dz-4\int \frac{z^{2}}{1+z^{3}}dz\)
\(=4\cdot\frac{z^{3}}{3} -\frac{4}{3}\ln|1+z^{3}|+C\)
\(=\frac{4}{3}z^{3} -\frac{4}{3}\ln|1+z^{3}|+C\)
\(=\frac{4}{3}x^{\frac{3}{4}} -\frac{4}{3}\ln\!\left(1+x^{\frac{3}{4}}\right)+C \quad (Ans)\)
Solution:
\(I=\int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx\)
\(=\int \frac{2\cos\frac{9x}{2}\cos\frac{x}{2}} {1-2\left(2\cos^{2}\frac{3x}{2}-1\right)}\,dx\)
\(=\int \frac{2\cos\frac{9x}{2}\cos\frac{x}{2}} {3-4\cos^{2}\frac{3x}{2}}\,dx\)
As, \(\cos 3\theta=4\cos^{3}\theta-3\cos\theta\)
\(\cos\frac{9x}{2} =4\cos^{3}\frac{3x}{2}-3\cos\frac{3x}{2}\)
So,
\(I=\int \frac{2\cos\frac{3x}{2}\left(4\cos^{2}\frac{3x}{2}-3\right) \cos\frac{x}{2}} {3-4\cos^{2}\frac{3x}{2}}\,dx\)
\(=-\int 2\cos\frac{3x}{2}\cos\frac{x}{2}\,dx\)
\(=-\int (\cos 2x+\cos x)\,dx\)
\(=-\frac{\sin 2x}{2}-\sin x+C\)
So, \(\displaystyle \int \frac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx =-\left(\frac{\sin 2x}{2}+\sin x\right)+C \quad (Answer)\)
Solution:
\(\frac{x^{2}}{(x^{2}+1)(x^{2}+2)} =\frac{A}{x^{2}+1}+\frac{B}{x^{2}+2}\)
\(\Rightarrow x^{2}=A(x^{2}+2)+B(x^{2}+1)\)
\(\Rightarrow x^{2}=(A+B)x^{2}+(2A+B)\)
So, \(A+B=1\) and \(2A+B=0\)
\(\Rightarrow A=-1,\; B=2\)
So, \(I=\int \left(\frac{-1}{x^{2}+1} +\frac{2}{x^{2}+2}\right)dx\)
\(=-\tan^{-1}x +\frac{2}{\sqrt2}\tan^{-1}\!\left(\frac{x}{\sqrt2}\right)+C\)
\(=-\tan^{-1}x +\sqrt2\,\tan^{-1}\!\left(\frac{x}{\sqrt2}\right)+C \quad (Ans)\)
Solution:
Let \(x+1=z^{2}\Rightarrow x=z^{2}-1\)
When \(x=8,\; z=3\) and when \(x=15,\; z=4\)
On differentiating w.r.t. \(x\), we get
\(dx=2z\,dz\)
So, \(I=\int_{3}^{4}\frac{2z\,dz}{(z^{2}-1-3)z}\)
\(=2\int_{3}^{4}\frac{dz}{z^{2}-4}\)
\(=2\cdot\frac{1}{4} \left[\ln\left|\frac{z-2}{z+2}\right|\right]_{3}^{4}\)
\(=\frac{1}{2} \left(\ln\frac{2}{6}-\ln\frac{1}{5}\right)\)
\(=\frac{1}{2}\ln\left(\frac{5}{3}\right) \quad (Ans)\)
Solution:
Let \(x=\sec\theta\)
\(dx=\sec\theta\tan\theta\,d\theta\)
When \(x=1,\;\theta=0\) and when \(x=2,\;\theta=\frac{\pi}{3}\)
Now, \(I=\int_{0}^{\pi/3} \frac{\sec\theta\tan\theta\,d\theta} {(1+\sec\theta)\tan\theta}\)
\(=\int_{0}^{\pi/3} \frac{\sec\theta}{1+\sec\theta}\,d\theta\)
\(=\int_{0}^{\pi/3} \frac{(1+\sec\theta)-1}{1+\sec\theta}\,d\theta\)
\(=\int_{0}^{\pi/3} d\theta -\int_{0}^{\pi/3}\frac{d\theta}{1+\sec\theta}\)
\(=\frac{\pi}{3} -\int_{0}^{\pi/3} \frac{\cos\theta}{1+\cos\theta}\,d\theta\)
\(=\frac{\pi}{3} -\int_{0}^{\pi/3} \frac{2\cos^{2}\frac{\theta}{2}-1} {2\cos^{2}\frac{\theta}{2}}\,d\theta\)
\(=\frac{\pi}{3} -\int_{0}^{\pi/3}d\theta +\frac{1}{2}\int_{0}^{\pi/3}\sec^{2}\frac{\theta}{2}\,d\theta\)
\(=\frac{\pi}{3}-\frac{\pi}{3} +\left[\tan\frac{\theta}{2}\right]_{0}^{\pi/3}\)
\(=\tan\frac{\pi}{6}-\tan0 =\frac{1}{\sqrt3}\quad (Ans)\)
Solution:
\(I=\int_{0}^{\pi/2}\frac{x\,dx}{\sin x+\cos x}\)
As, \(\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx\)
\(I=\int_{0}^{\pi/2} \frac{\left(\frac{\pi}{2}-x\right)dx} {\sin\left(\frac{\pi}{2}-x\right) +\cos\left(\frac{\pi}{2}-x\right)}\)
\(=\int_{0}^{\pi/2} \frac{\left(\frac{\pi}{2}-x\right)dx} {\cos x+\sin x}\)
\(=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{dx}{\sin x+\cos x}-I\)
\(\Rightarrow 2I= \frac{\pi}{2}\int_{0}^{\pi/2} \frac{dx}{\sin x+\cos x}\)
\(\Rightarrow I= \frac{\pi}{4}\int_{0}^{\pi/2} \frac{dx}{\sin x+\cos x}\)
\(=\frac{\pi}{4\sqrt2} \int_{0}^{\pi/2} \text{cosec}\!\left(x+\frac{\pi}{4}\right)dx\)
\(=\frac{\pi}{4\sqrt2} \left[\ln\left|\text{cosec}\!\left(x+\frac{\pi}{4}\right) -\cot\!\left(x+\frac{\pi}{4}\right)\right| \right]_{0}^{\pi/2}\)
\(=\frac{\pi}{4\sqrt2} \ln\left(\frac{\sqrt2+1}{\sqrt2-1}\right)\)
\(=\frac{\pi}{2\sqrt2} \ln(\sqrt2+1)\quad (Ans)\)
Solution:
\(I=\int_{0}^{\pi} \frac{x\,dx}{a^{2}\cos^{2}x+b^{2}\sin^{2}x}\)
As, \(\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx\)
\(I=\int_{0}^{\pi} \frac{(\pi-x)dx} {a^{2}\cos^{2}(\pi-x)+b^{2}\sin^{2}(\pi-x)}\)
\(=\int_{0}^{\pi} \frac{(\pi-x)dx} {a^{2}\cos^{2}x+b^{2}\sin^{2}x}\)
\(=\pi\int_{0}^{\pi} \frac{dx}{a^{2}\cos^{2}x+b^{2}\sin^{2}x}-I\)
\(\Rightarrow 2I= \pi\int_{0}^{\pi} \frac{dx}{a^{2}\cos^{2}x+b^{2}\sin^{2}x}\)
\(=\pi\int_{0}^{\pi} \frac{\sec^{2}x\,dx}{a^{2}+b^{2}\tan^{2}x}\)
As, \(\int_{0}^{\pi}f(x)dx =2\int_{0}^{\pi/2}f(x)dx\) when \(f(\pi-x)=f(x)\)
\(2I= 2\pi\int_{0}^{\pi/2} \frac{\sec^{2}x\,dx}{a^{2}+b^{2}\tan^{2}x}\)
\(\Rightarrow I= \pi\int_{0}^{\pi/2} \frac{\sec^{2}x\,dx}{a^{2}+b^{2}\tan^{2}x}\)
Let \(\tan x=z\Rightarrow \sec^{2}x\,dx=dz\)
When \(x=0,\;z=0\) and when \(x=\frac{\pi}{2},\;z=\infty\)
\(I= \frac{\pi}{b^{2}} \int_{0}^{\infty} \frac{dz}{\left(\frac{a}{b}\right)^{2}+z^{2}}\)
\(=\frac{\pi}{b^{2}} \cdot\frac{b}{a} \left[\tan^{-1}\!\left(\frac{z}{a/b}\right)\right]_{0}^{\infty}\)
\(=\frac{\pi}{ab} \left(\frac{\pi}{2}\right)\)
\(=\frac{\pi^{2}}{2ab}\quad (Ans)\)
Solution:
As, \(\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx\)
So, \(I=\int_{2}^{3} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{5-x}}\,dx\)
\(=\int_{2}^{3} \frac{\sqrt{5-x}}{\sqrt{5-x}+\sqrt{x}}\,dx\)
Now, \(2I=\int_{2}^{3} \frac{\sqrt{x}+\sqrt{5-x}} {\sqrt{x}+\sqrt{5-x}}\,dx\)
\(\Rightarrow 2I=\int_{2}^{3}dx\)
\(\Rightarrow I=\frac{1}{2}[x]_{2}^{3}\)
\(=\frac{1}{2}\quad (Ans)\)
Solution:
\(I=\int \frac{x^{4}+1}{x^{6}+1}\,dx\)
\(=\int \frac{(x^{4}-x^{2}+1)+x^{2}}{(x^{2})^{3}+1}\,dx\)
\(=\int \frac{x^{4}-x^{2}+1}{(x^{2}+1)(x^{4}-x^{2}+1)}\,dx +\int \frac{x^{2}}{x^{6}+1}\,dx\)
\(=\int \frac{dx}{1+x^{2}} +\frac{1}{3}\int \frac{3x^{2}\,dx}{(x^{3})^{2}+1}\)
Let \(x^{3}=z\Rightarrow 3x^{2}dx=dz\)
So, \(I=\tan^{-1}x +\frac{1}{3}\int \frac{dz}{1+z^{2}}+C\)
\(=\tan^{-1}x +\frac{1}{3}\tan^{-1}z+C\)
\(=\tan^{-1}x +\frac{1}{3}\tan^{-1}(x^{3})+C \quad (Ans)\)
Solution:
\(I=\int \sqrt{1+\sec x}\,dx\)
\(=\int \sqrt{\frac{1+\cos x}{\cos x}}\,dx\)
\(=\int \sqrt{\frac{2\cos^{2}\frac{x}{2}} {1-2\sin^{2}\frac{x}{2}}}\,dx\)
\(=\int \frac{\sqrt2\cos\frac{x}{2}} {\sqrt{1-2\sin^{2}\frac{x}{2}}}\,dx\)
Let \(\sin\frac{x}{2}=z\)
\(\frac{1}{2}\cos\frac{x}{2}\,dx=dz \Rightarrow \cos\frac{x}{2}\,dx=2dz\)
So, \(I=\int \frac{\sqrt2\cdot2\,dz} {\sqrt{1-2z^{2}}}\)
\(=2\int \frac{dz} {\sqrt{\left(\frac{1}{\sqrt2}\right)^{2}-z^{2}}}\)
\(=2\sin^{-1}\!\left( \frac{z}{1/\sqrt2}\right)+C\)
\(=2\sin^{-1}\!\left( \sqrt2\sin\frac{x}{2}\right)+C \quad (Ans)\)
Solution:
\(\sin x=\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}} =\frac{2\tan\frac{x}{2}}{\sec^{2}\frac{x}{2}}\)
Now, \(I=\int_{0}^{\pi/2} \frac{dx}{5+\frac{8\tan\frac{x}{2}} {\sec^{2}\frac{x}{2}}}\)
\(=\int_{0}^{\pi/2} \frac{\sec^{2}\frac{x}{2}\,dx} {5\sec^{2}\frac{x}{2}+8\tan\frac{x}{2}}\)
\(=\int_{0}^{\pi/2} \frac{\sec^{2}\frac{x}{2}\,dx} {5+5\tan^{2}\frac{x}{2}+8\tan\frac{x}{2}}\)
Let \(\tan\frac{x}{2}=z\)
\(\frac{1}{2}\sec^{2}\frac{x}{2}\,dx=dz \Rightarrow \sec^{2}\frac{x}{2}\,dx=2dz\)
When \(x=0,\;z=0\) and when \(x=\frac{\pi}{2},\;z=1\)
So, \(I=\int_{0}^{1} \frac{2\,dz}{5z^{2}+8z+5}\)
\(5z^{2}+8z+5 =5\left(z+\frac{4}{5}\right)^{2} +\left(\frac{3}{5}\right)^{2}\)
\(I=\frac{2}{5} \int_{0}^{1} \frac{dz}{\left(z+\frac{4}{5}\right)^{2} +\left(\frac{3}{5}\right)^{2}}\)
\(=\frac{2}{5}\cdot \frac{5}{3} \left[\tan^{-1} \!\left(\frac{z+\frac{4}{5}} {\frac{3}{5}}\right)\right]_{0}^{1}\)
\(=\frac{2}{3} \left(\tan^{-1}(3)-\tan^{-1}\!\left(\frac{4}{3}\right)\right)\)
\(=\frac{2}{3} \tan^{-1}\!\left(\frac{1}{3}\right)\)
Hence, \(\displaystyle \int_{0}^{\pi/2}\frac{dx}{5+4\sin x} =\frac{2}{3}\tan^{-1}\!\left(\frac{1}{3}\right) \quad (Answer)\)
Solution:
Let, \(\cos x=z\)
Differentiating w.r.t. \(x\), we get
\(\sin x\,dx=-dz\)
\[ \begin{array}{c|cc} x & 0 & \pi \\ \hline z & 1 & -1 \end{array} \]
So, \(I=\int_{1}^{-1}\frac{-dz}{1+z^{2}}\)
\(=-\left[\tan^{-1}(z)\right]_{-1}^{1}\)
\(=-(\tan^{-1}(1)-\tan^{-1}(-1))\)
\(=-(\frac{\pi}{4}+\frac{\pi}{4}) \quad [\because \tan^{-1}(-x)=-\tan^{-1}x]\)
\(=-\frac{\pi}{2}\)
Hence, \(\displaystyle \int_{0}^{\pi} \frac{\sin x\,dx}{1+\cos^{2}x} =-\frac{\pi}{2}\quad (Answer)\)
Solution:
Given integral
\(\displaystyle \int \frac{\sec x}{1+\text{cosec} x}\,dx\)
\(=\int \frac{\sin x}{\cos x(1+\sin x)}\,dx\)
\(=\int \frac{\sin x\cos x}{(1+\sin x)^{2}(1-\sin x)}\,dx\)
\(=\int \frac{t}{(1+t)^{2}(1-t)}\,dt \qquad [\sin x=t,\; \cos x\,dx=dt]\)
Let, \[ \frac{t}{(1+t)^{2}(1-t)} =\frac{A}{1+t} +\frac{B}{(1+t)^{2}} +\frac{C}{1-t} \]
\(t=A(1+t)(1-t)+B(1-t)+C(1+t)^{2}\) (an identity)
Put \(t=-1\Rightarrow -1=2B,\;B=-\frac12\)
Put \(t=1\Rightarrow 1=4C,\;C=\frac14\)
Put \(t=0\Rightarrow 0=A+B+C\Rightarrow A=\frac14\)
Therefore the required integral
\(=\frac14\int\frac{1}{1+t}\,dt -\frac12\int\frac{1}{(1+t)^{2}}\,dt +\frac14\int\frac{1}{1-t}\,dt\)
\(=\frac14\log|1+t| +\frac12\frac{1}{1+t} -\frac14\log|1-t|+C\)
\(=\frac14\log|1+\sin x| +\frac12\frac{1}{1+\sin x} -\frac14\log|1-\sin x|+C\)
\(=\frac14\log\left|\frac{1+\sin x}{1-\sin x}\right| +\frac12\frac{1}{1+\sin x}+C\)
📙 4 Mark Integrals
15+ problemsSolution:
Let, \(x=a\tan^{2}\theta\)
Differentiating w.r.t. \(x\), we get
\(dx=2a\tan\theta\sec^{2}\theta\,d\theta\)
Now, \(I=\int \cos^{-1} \sqrt{\frac{a\tan^{2}\theta}{a+a\tan^{2}\theta}} \cdot 2a\tan\theta\sec^{2}\theta\,d\theta\)
\(=\int \cos^{-1}\!\left(\frac{\tan\theta}{\sec\theta}\right) (2a\tan\theta\sec^{2}\theta)\,d\theta\)
\(=2a\int \cos^{-1}(\sin\theta) (\tan\theta\sec^{2}\theta)\,d\theta\)
\(=2a\int \cos^{-1} \left(\cos\left(\frac{\pi}{2}-\theta\right)\right) \tan\theta\sec^{2}\theta\,d\theta\)
\(=2a\int \left(\frac{\pi}{2}-\theta\right) \tan\theta\sec^{2}\theta\,d\theta\)
\(=a\pi\int \tan\theta\sec^{2}\theta\,d\theta -2a\int \theta(\tan\theta\sec^{2}\theta)\,d\theta\)
\(=a\pi\int \tan\theta\,d(\tan\theta) -2a\left[\theta\frac{\tan^{2}\theta}{2} -\int\frac{\tan^{2}\theta}{2}\,d\theta\right]\)
\(=\frac{a\pi\tan^{2}\theta}{2} -a\theta\tan^{2}\theta +\frac{a}{2}\int (\sec^{2}\theta-1)\,d\theta +C\)
\(=\frac{a\pi\tan^{2}\theta}{2} -a\theta\tan^{2}\theta +\frac{a}{2}\tan\theta -\frac{a}{2}\theta +C\)
Substituting back, \(\tan^{2}\theta=\frac{x}{a}\), \(\tan\theta=\sqrt{\frac{x}{a}}\), \(\theta=\tan^{-1}\sqrt{\frac{x}{a}}\)
\(=\frac{\pi x}{2} -x\tan^{-1}\!\sqrt{\frac{x}{a}} +\frac{\sqrt{x}}{2\sqrt{a}} -\frac{1}{2}\tan^{-1}\!\sqrt{\frac{x}{a}}+C\quad (Ans)\)
Solution:
\(I=\int \frac{x^{2}\,dx}{(x\sin x+\cos x)^{2}}\)
\(=\int \frac{x}{\cos x}\cdot \frac{x\cos x}{(x\sin x+\cos x)^{2}}\,dx\)
Let, \(\displaystyle \frac{1}{x\sin x+\cos x}=z\)
Differentiating w.r.t. \(x\), we get
\(\displaystyle -\frac{x\cos x}{(x\sin x+\cos x)^{2}}\,dx=dz\)
\(\displaystyle \frac{x\cos x}{(x\sin x+\cos x)^{2}}\,dx=-dz\)
So, \(I_{2}=\int \frac{x\cos x}{(x\sin x+\cos x)^{2}}\,dx =-\int dz=-z =-\frac{1}{x\sin x+\cos x}\)
Now,
\(I=\int \frac{x}{\cos x}\cdot \frac{x\cos x}{(x\sin x+\cos x)^{2}}\,dx\)
\(=\frac{x}{\cos x}(I_{2}) -\int \left[\frac{d}{dx}\left(\frac{x}{\cos x}\right)\right] (I_{2})\,dx +C\)
\(=-\frac{x}{\cos x(x\sin x+\cos x)} +\int \frac{x\sin x+\cos x}{\cos^{2}x} \cdot\frac{1}{x\sin x+\cos x}\,dx +C\)
\(=-\frac{x}{\cos x(x\sin x+\cos x)} +\int \sec^{2}x\,dx +C\)
\(=-\frac{x}{\cos x(x\sin x+\cos x)} +\tan x +C\)
Hence, \(\displaystyle \int \frac{x^{2}\,dx}{(x\sin x+\cos x)^{2}} =-\frac{x}{\cos x(x\sin x+\cos x)} +\tan x +C\quad (Answer)\)
Solution:
\(I=\int \frac{dx}{3+2\sin x+\cos x}\)
\(=\int \frac{dx} {3\sin^{2}\frac{x}{2} +3\cos^{2}\frac{x}{2} +2\cdot2\sin\frac{x}{2}\cos\frac{x}{2} +\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}\)
\(=\int \frac{dx} {2\sin^{2}\frac{x}{2} +4\cos^{2}\frac{x}{2} +4\sin\frac{x}{2}\cos\frac{x}{2}}\)
\(=\int \frac{\sec^{2}\frac{x}{2}\,dx} {2\tan^{2}\frac{x}{2} +4+4\tan\frac{x}{2}}\)
\(=\int \frac{\frac12\sec^{2}\frac{x}{2}\,dx} {\tan^{2}\frac{x}{2} +2\tan\frac{x}{2} +2}\)
Let, \(\tan\frac{x}{2}=z\)
\(\frac12\sec^{2}\frac{x}{2}\,dx=dz\)
\(I=\int \frac{dz}{z^{2}+2z+2}\)
\(=\int \frac{dz}{(z+1)^{2}+1}\)
\(=\tan^{-1}(z+1)+C\)
\(=\tan^{-1}\!\left(1+\tan\frac{x}{2}\right)+C \quad (Answer)\)
Solution:
\(5\cos x+4\sin x =m(4\cos x+5\sin x) +n\frac{d}{dx}(4\cos x+5\sin x)\)
\(=4m\cos x+5m\sin x +n(-4\sin x+5\cos x)\)
\(=(4m+5n)\cos x+(5m-4n)\sin x\)
So, \(4m+5n=5\)
\(5m-4n=4\)
\(m=\frac{5-5n}{4}\)
\(5\left(\frac{5-5n}{4}\right)-4n=4\)
\(25-25n-16n=16\)
\(9=41n\Rightarrow n=\frac{9}{41}\)
\(m=\frac{5-\frac{45}{41}}{4} =\frac{160}{4\times41} =\frac{40}{41}\)
Now,
\(I=\int \frac{m(4\cos x+5\sin x) +n\frac{d}{dx}(4\cos x+5\sin x)} {4\cos x+5\sin x}\,dx\)
\(=m\int dx +n\int \frac{d(4\cos x+5\sin x)} {4\cos x+5\sin x}\)
\(=mx+n\ln|4\cos x+5\sin x|+C\)
\(=\frac{40}{41}x +\frac{9}{41}\ln|4\cos x+5\sin x| +C\quad (Answer)\)
Solution:
\(I=\int \frac{dx}{1+\cos\alpha\cos x}\)
\(=\int \frac{dx} {1+\cos\alpha\left(\frac{1-\tan^{2}\frac{x}{2}} {1+\tan^{2}\frac{x}{2}}\right)}\)
\(=\int \frac{1+\tan^{2}\frac{x}{2}\,dx} {1+\tan^{2}\frac{x}{2} +\cos\alpha(1-\tan^{2}\frac{x}{2})}\)
\(=\int \frac{\sec^{2}\frac{x}{2}\,dx} {(1+\cos\alpha) +(1-\cos\alpha)\tan^{2}\frac{x}{2}}\)
\(=\frac{2}{1-\cos\alpha} \int \frac{\frac12\sec^{2}\frac{x}{2}\,dx} {\left(\cot\frac{\alpha}{2}\right)^{2} +\tan^{2}\frac{x}{2}}\)
Let, \(\tan\frac{x}{2}=z\)
\(\frac12\sec^{2}\frac{x}{2}\,dx=dz\)
\(I=\frac{2}{1-\cos\alpha} \int \frac{dz} {\left(\cot\frac{\alpha}{2}\right)^{2}+z^{2}}\)
\(=\frac{2}{1-\cos\alpha} \cdot\frac{1}{\cot\frac{\alpha}{2}} \tan^{-1}\!\left(\frac{z} {\cot\frac{\alpha}{2}}\right)+C\)
\(=\frac{2}{1-\cos\alpha} \tan\frac{\alpha}{2} \tan^{-1}\!\left( \tan\frac{x}{2}\tan\frac{\alpha}{2}\right)+C\)
\(=\frac{2}{1+\cos\alpha} \tan^{-1}\!\left( \tan\frac{x}{2}\tan\frac{\alpha}{2}\right)+C \quad (Answer)\)
Solution:
\(I=\int \frac{x}{(1+x)(1+x^{2})}\,dx\)
\(=\int \left(\frac{A}{1+x} +\frac{Bx+C}{1+x^{2}}\right)dx\)
\[ \frac{x}{(1+x)(1+x^{2})} =\frac{A}{1+x} +\frac{Bx+C}{1+x^{2}} \]
\(x=A(1+x^{2})+(Bx+C)(1+x)\)
Put \(x=-1\Rightarrow -1=2A\Rightarrow A=-\frac12\)
Put \(x=0\Rightarrow 0=A+C\Rightarrow C=\frac12\)
Put \(x=1\Rightarrow 1=2A+2(B+C)\)
\(1=2\left(-\frac12\right)+2\left(B+\frac12\right)\)
\(1=-1+2B+1\Rightarrow B=\frac12\)
So, \(A=-\frac12,\;B=\frac12,\;C=\frac12\)
\(I=\int \frac{-1/2}{1+x}\,dx +\int \frac{\frac12 x+\frac12}{1+x^{2}}\,dx\)
\(=-\frac12\ln|1+x| +\frac12\int \frac{x}{1+x^{2}}\,dx +\frac12\int \frac{dx}{1+x^{2}}\)
\(=-\frac12\ln|1+x| +\frac14\ln|1+x^{2}| +\frac12\tan^{-1}x+C\)
\(=\boxed{ -\frac12\ln|1+x| +\frac14\ln|1+x^{2}| +\frac12\tan^{-1}x +C}\quad (Answer)\)
Solution:
\(I=\int \frac{dx}{x^{3}+1}\)
\(=\int \frac{dx}{(x+1)(x^{2}-x+1)}\)
\(\displaystyle \frac{1}{(x+1)(x^{2}-x+1)} =\frac{A}{x+1} +\frac{Bx+C}{x^{2}-x+1}\)
\(1=A(x^{2}-x+1)+(Bx+C)(x+1)\)
Put \(x=-1\Rightarrow 1=3A\Rightarrow A=\frac13\)
Put \(x=0\Rightarrow 1=A+C\Rightarrow C=\frac23\)
Put \(x=1\Rightarrow 1=A+2(B+C)\Rightarrow B=-\frac13\)
So, \(A=\frac13,\; B=-\frac13,\; C=\frac23\)
\(I=\frac13\int \frac{dx}{x+1} +\int \frac{-\frac13 x+\frac23}{x^{2}-x+1}\,dx\)
\(=\frac13\ln|x+1| -\frac13\int \frac{x-2}{x^{2}-x+1}\,dx\)
\(=\frac13\ln|x+1| -\frac16\int \frac{2x-1}{x^{2}-x+1}\,dx +\frac12\int \frac{dx}{x^{2}-x+1}\)
\(=\frac13\ln|x+1| -\frac16\ln|x^{2}-x+1| +\frac12\int \frac{dx}{(x-\frac12)^{2} +(\frac{\sqrt3}{2})^{2}}\)
\(=\frac13\ln|x+1| -\frac16\ln|x^{2}-x+1| +\frac{1}{\sqrt3} \tan^{-1}\!\left( \frac{2x-1}{\sqrt3}\right)+C\)
\(\boxed{ \frac13\ln|x+1| -\frac16\ln|x^{2}-x+1| +\frac{1}{\sqrt3} \tan^{-1}\!\left( \frac{2x-1}{\sqrt3}\right) +C}\)
Solution:
\( x^{4}+1 =(x^{2}+\sqrt{2}x+1)(x^{2}-\sqrt{2}x+1) \)
\( \frac{1}{x^{4}+1} = \frac{Ax+B}{x^{2}+\sqrt{2}x+1} + \frac{Cx+D}{x^{2}-\sqrt{2}x+1} \)
Solving,
\( A=\frac{1}{2\sqrt{2}}, \; B=\frac{1}{2}, \; C=-\frac{1}{2\sqrt{2}}, \; D=\frac{1}{2} \)
Hence,
\( \int\frac{dx}{x^{4}+1} = \frac{1}{2\sqrt{2}} \int \frac{x\,dx}{x^{2}+\sqrt{2}x+1} + \frac{1}{2} \int \frac{dx}{x^{2}+\sqrt{2}x+1} -\frac{1}{2\sqrt{2}} \int \frac{x\,dx}{x^{2}-\sqrt{2}x+1} + \frac{1}{2} \int \frac{dx}{x^{2}-\sqrt{2}x+1} \)
Completing the square,
\( x^{2}\pm\sqrt{2}x+1 = \left(x\pm\frac{1}{\sqrt{2}}\right)^{2} +\frac{1}{2} \)
Integrating,
\( \int\frac{dx}{x^{4}+1} = \frac{1}{2\sqrt{2}} \ln\left| \frac{x^{2}+\sqrt{2}x+1} {x^{2}-\sqrt{2}x+1} \right| + \frac{1}{\sqrt{2}} \tan^{-1} \frac{\sqrt{2}x}{1-x^{2}} +C \)
Solution:
\(\displaystyle \lim_{n\to\infty} \sum_{r=1}^{n} \frac{n+r}{n^{2}+r^{2}}\)
\(=\lim_{n\to\infty} \frac1n \sum_{r=1}^{n} \frac{n^{2}+nr}{n^{2}+r^{2}}\)
\(=\lim_{n\to\infty} \frac1n \sum_{r=1}^{n} \frac{1+\frac{r}{n}} {1+\left(\frac{r}{n}\right)^{2}}\)
\(=\int_{0}^{1} \frac{1+x}{1+x^{2}}\,dx\)
\(=\int_{0}^{1} \frac{dx}{1+x^{2}} +\int_{0}^{1} \frac{x\,dx}{1+x^{2}}\)
\(=\left[\tan^{-1}x\right]_{0}^{1} +\frac12\left[\ln(1+x^{2})\right]_{0}^{1}\)
\(=\tan^{-1}(1)-\tan^{-1}(0) +\frac12(\ln2-\ln1)\)
\(=\frac{\pi}{4} +\frac12\ln2\)
\(\boxed{ \frac{\pi}{4} +\frac12\ln2}\)
Solution:
Let, \(A=\left(\frac{n!}{n^{n}}\right)^{\frac1n}\)
\(\Rightarrow \log_e A =\frac1n\log_e\!\left(\frac{1\cdot2\cdot3\cdots n} {n\cdot n\cdot n\cdots n}\right)\)
\(=\frac1n\log_e\!\left( \frac1n\cdot\frac2n\cdot\frac3n\cdots\frac{n}{n} \right)\)
\(=\frac1n\sum_{r=1}^{n} \log_e\!\left(\frac{r}{n}\right)\)
Now,
\(\displaystyle \lim_{n\to\infty}\log_e A =\lim_{n\to\infty} \frac1n\sum_{r=1}^{n} \log_e\!\left(\frac{r}{n}\right)\)
\(=\int_{0}^{1}\log_e x\,dx\)
\(=\left[x\log_e x-x\right]_{0}^{1}\)
\(=(1\cdot0-1)-(0-0)=-1\)
\(\Rightarrow \log_e\!\left( \lim_{n\to\infty} \left(\frac{n!}{n^{n}}\right)^{\!\frac1n} \right)=-1\)
\(\therefore \displaystyle \lim_{n\to\infty} \left(\frac{n!}{n^{n}}\right)^{\!\frac1n} =e^{-1} =\frac1e\quad (\text{Proved})\)
Solution:
As, \(\displaystyle \int_{0}^{a}f(x)\,dx =\int_{0}^{a}f(a-x)\,dx\)
Let \(I=\displaystyle \int_{0}^{\pi} \frac{x\,dx} {(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}\)
Replacing \(x\) by \(\pi-x\),
\(I=\displaystyle \int_{0}^{\pi} \frac{\pi-x} {(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}\,dx\)
Adding,
\(2I=\pi \displaystyle\int_{0}^{\pi} \frac{dx} {(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}\)
\(\displaystyle \int_{0}^{\pi} f(x)\,dx =2\int_{0}^{\pi/2} f(x)\,dx \quad\text{when }f(\pi-x)=f(x)\)
Hence,
\(2I=\pi\cdot 2\int_{0}^{\pi/2} \frac{dx} {(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}\)
\(\Rightarrow I=\pi \int_{0}^{\pi/2} \frac{dx} {(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}\)
Put \(b\tan x=a\tan\theta\)
Then \(b\sec^{2}x\,dx =a\sec^{2}\theta\,d\theta\)
After substitution,
\(I=\frac{\pi}{a^{3}b^{3}} \int_{0}^{\pi/2} \left(b^{2}\cos^{2}\theta +a^{2}\sin^{2}\theta\right)d\theta\)
\(=\frac{\pi}{a^{3}b^{3}} \left[ b^{2}\int_{0}^{\pi/2}\cos^{2}\theta\,d\theta +a^{2}\int_{0}^{\pi/2}\sin^{2}\theta\,d\theta \right]\)
\(=\frac{\pi}{a^{3}b^{3}} \left[ b^{2}\frac{\pi}{4} +a^{2}\frac{\pi}{4} \right]\)
\(=\frac{\pi^{2}(a^{2}+b^{2})} {4a^{3}b^{3}}\quad(\text{Proved})\)
Solution:
Let, \(\sin\theta-\cos\theta=z\)
\(\Rightarrow (\sin\theta+\cos\theta)d\theta=dz\)
\(d\theta=\dfrac{dz}{\sin\theta+\cos\theta}\)
\(\sin2\theta=2\sin\theta\cos\theta\)
\(=1-(\sin^{2}\theta+\cos^{2}\theta-2\sin\theta\cos\theta)\)
\(=1-(\sin\theta-\cos\theta)^{2}\)
\(=1-z^{2}\)
When \(\theta=0, z=-1\) and when \(\theta=\frac{\pi}{4}, z=0\)
\( I=\int_{-1}^{0} \frac{dz}{9+16(1-z^{2})} \)
\( =\int_{-1}^{0} \frac{dz}{25-16z^{2}} \)
\( =\frac{1}{16} \int_{-1}^{0} \frac{dz}{\left(\frac{5}{4}\right)^{2}-z^{2}} \)
\( =\frac{1}{16}\cdot\frac{1}{2\left(\frac{5}{4}\right)} \left[ \ln\left| \frac{\frac{5}{4}+z} {\frac{5}{4}-z} \right| \right]_{-1}^{0} \)
\( =\frac{1}{40} \left[ \ln\left| \frac{\frac{5}{4}+0} {\frac{5}{4}-0} \right| -\ln\left| \frac{\frac{5}{4}-1} {\frac{5}{4}+1} \right| \right] \)
\( =\frac{1}{40} \left[ 0-\ln\left| \frac{1}{9} \right| \right] \)
\( =\frac{1}{40}\log_e 9 \)
\( =\frac{1}{20}\log_e 3\quad(\text{Ans}) \)
Solution:
Let, \(\cos x = z\)
\(-\sin x\,dx = dz\)
\(\Rightarrow \sin x\,dx = -dz\)
When \(x=0, z=1\) and when \(x=\frac{\pi}{2}, z=0\)
\( I=\int_{1}^{0} \frac{-dz} {(a+z)(b+z)} \)
\( =\int_{0}^{1} \frac{dz} {(a+z)(b+z)} \)
\( =\int_{0}^{1} \frac{dz} {z^{2}+(a+b)z+ab} \)
\( =\int_{0}^{1} \frac{dz} {z^{2}+2z\frac{a+b}{2}+\left(\frac{a+b}{2}\right)^{2} +ab-\left(\frac{a+b}{2}\right)^{2}} \)
\( =\int_{0}^{1} \frac{dz} {\left(z+\frac{a+b}{2}\right)^{2} -\left(\frac{a-b}{2}\right)^{2}} \)
\( =\frac{1}{2\left(\frac{a-b}{2}\right)} \left[ \ln\left| \frac{z+\frac{a+b}{2}-\frac{a-b}{2}} {z+\frac{a+b}{2}+\frac{a-b}{2}} \right| \right]_{0}^{1} \)
\( =\frac{1}{a-b} \left[ \ln\left| \frac{2z+2b} {2z+2a} \right| \right]_{0}^{1} \)
\( =\frac{1}{a-b} \left[ \ln\left| \frac{1+b}{1+a} \right| -\ln\left| \frac{b}{a} \right| \right] \)
\( =\frac{1}{a-b} \ln\left( \frac{a}{b}\cdot \frac{1+b}{1+a} \right) \)
Solution:
LHS: \( \lim_{n\to\infty} \left[ \left(1+\frac{1^{2}}{n^{2}}\right)^{\frac{2}{n^{2}}} \left(1+\frac{2^{2}}{n^{2}}\right)^{\frac{4}{n^{2}}} \left(1+\frac{3^{2}}{n^{2}}\right)^{\frac{6}{n^{2}}} \cdots \left(1+\frac{n^{2}}{n^{2}}\right)^{\frac{2n}{n^{2}}} \right] \)
Let, \( A= \left(1+\frac{1^{2}}{n^{2}}\right)^{\frac{2}{n^{2}}} \left(1+\frac{2^{2}}{n^{2}}\right)^{\frac{4}{n^{2}}} \left(1+\frac{3^{2}}{n^{2}}\right)^{\frac{6}{n^{2}}} \cdots \left(1+\frac{n^{2}}{n^{2}}\right)^{\frac{2n}{n^{2}}} \)
\( \log_e A= \frac{2}{n^{2}}\log_e\left(1+\frac{1^{2}}{n^{2}}\right) +\frac{4}{n^{2}}\log_e\left(1+\frac{2^{2}}{n^{2}}\right) +\frac{6}{n^{2}}\log_e\left(1+\frac{3^{2}}{n^{2}}\right) +\cdots +\frac{2n}{n^{2}}\log_e\left(1+\frac{n^{2}}{n^{2}}\right) \)
\( =\frac{2}{n^{2}} \left[ 1\log_e\left(1+\frac{1^{2}}{n^{2}}\right) +2\log_e\left(1+\frac{2^{2}}{n^{2}}\right) +\cdots +n\log_e\left(1+\frac{n^{2}}{n^{2}}\right) \right] \)
\( =\frac{2}{n} \sum_{r=1}^{n} \frac{r}{n} \log_e\left(1+\left(\frac{r}{n}\right)^{2}\right) \)
\( \lim_{n\to\infty}\log_e A = \lim_{n\to\infty} \frac{2}{n} \sum_{r=1}^{n} \frac{r}{n} \log_e\left(1+\left(\frac{r}{n}\right)^{2}\right) \)
\( =2\int_{0}^{1} x\log_e(1+x^{2})\,dx \)
Let \(I=\displaystyle\int_{0}^{1}x\log_e(1+x^{2})\,dx\)
Let \(1+x^{2}=z\)
Then \(2x\,dx=dz\)
\(x\,dx=\frac{dz}{2}\)
When \(x=0, z=1\) and when \(x=1, z=2\)
\( I=\int_{1}^{2}\log_e z\frac{dz}{2} \)
\( =\frac{1}{2}\int_{1}^{2}\log_e z\,dz \)
\( =\frac{1}{2}\left[z\log_e z-z\right]_{1}^{2} \)
\( =\frac{1}{2}\{(2\log_e2-2)-(1\log_e1-1)\} \)
\( =\frac{1}{2}(2\log_e2-1) \)
\( =\log_e2-\frac{1}{2} \)
Hence, \( \log_e(\lim_{n\to\infty}A) =2\left(\log_e2-\frac{1}{2}\right) \)
\( =2\log_e2-1 \)
\( \lim_{n\to\infty}A =e^{2\log_e2-1} \)
\( =\frac{e^{2\log_e2}}{e} \)
\( =\frac{4}{e}\quad(\text{Proved}) \)
Solution:
Let \(x=\tan\theta\)
\(dx=\sec^{2}\theta\,d\theta\)
When \(x=0, \theta=0\) and when \(x=1, \theta=\frac{\pi}{4}\)
\( I=\int_{0}^{\pi/4} \frac{\tan^{-1}(\tan\theta)\sec^{2}\theta\,d\theta} {(1+\tan^{2}\theta)^{3/2}} \)
\( =\int_{0}^{\pi/4} \frac{\theta\sec^{2}\theta\,d\theta} {(\sec^{2}\theta)^{3/2}} \)
\( =\int_{0}^{\pi/4} \frac{\theta\sec^{2}\theta\,d\theta} {\sec^{3}\theta} \)
\( =\int_{0}^{\pi/4} \theta\cos\theta\,d\theta \)
\( I=\left[ \theta\sin\theta -\int \sin\theta\,d\theta \right]_{0}^{\pi/4} \)
\( =\left[ \theta\sin\theta +\cos\theta \right]_{0}^{\pi/4} \)
\( =\frac{\pi}{4}\cdot\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}} -1 \)
\( =\frac{\pi}{4\sqrt{2}} +\frac{1}{\sqrt{2}} -1\quad(\text{Ans}) \)
Solution:
\(I=\displaystyle\int_{0}^{\pi/2} \frac{x\sin x\cos x\,dx} {(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}}\)
Let \(a^{2}\cos^{2}x+b^{2}\sin^{2}x=z\)
Differentiating w.r.t. \(x\),
\( (2b^{2}\sin x\cos x-2a^{2}\sin x\cos x)dx=dz \)
\( \sin x\cos x\,dx= \frac{dz}{2(b^{2}-a^{2})} \)
\( \int \frac{\sin x\cos x\,dx} {(a^{2}\cos^{2}x+b^{2}\sin^{2}x)^{2}} = \int \frac{\frac{dz}{2(b^{2}-a^{2})}} {z^{2}} \)
\( =\frac{1}{2(b^{2}-a^{2})} \int z^{-2}dz \)
\( =\frac{1}{2(b^{2}-a^{2})} \left(-\frac{1}{z}\right) \)
\( =\frac{1}{2(a^{2}-b^{2})} \frac{1} {a^{2}\cos^{2}x+b^{2}\sin^{2}x} \)
Using integration by parts,
\( I= \left[ \frac{x}{2(a^{2}-b^{2})} \frac{1} {a^{2}\cos^{2}x+b^{2}\sin^{2}x} \right]_{0}^{\pi/2} - \int_{0}^{\pi/2} \frac{1}{2(a^{2}-b^{2})} \frac{dx} {a^{2}\cos^{2}x+b^{2}\sin^{2}x} \)
\( = \frac{\pi}{4(a^{2}-b^{2})} \frac{1}{b^{2}} - \frac{1}{2(a^{2}-b^{2})} \int_{0}^{\pi/2} \frac{dx} {a^{2}\cos^{2}x+b^{2}\sin^{2}x} \)
\( = \frac{\pi}{4b^{2}(a^{2}-b^{2})} - \frac{1}{2(a^{2}-b^{2})} \int_{0}^{\pi/2} \frac{\sec^{2}x\,dx} {a^{2}+b^{2}\tan^{2}x} \)
Let \(\tan x=z\), \(\sec^{2}x\,dx=dz\)
When \(x=0, z=0\) and when \(x=\frac{\pi}{2}, z=\infty\)
\( I= \frac{\pi}{4b^{2}(a^{2}-b^{2})} - \frac{1}{2(a^{2}-b^{2})} \int_{0}^{\infty} \frac{dz} {a^{2}+b^{2}z^{2}} \)
\( = \frac{\pi}{4b^{2}(a^{2}-b^{2})} - \frac{1}{2b^{2}(a^{2}-b^{2})} \int_{0}^{\infty} \frac{dz} {\left(\frac{a}{b}\right)^{2}+z^{2}} \)
\( = \frac{\pi}{4b^{2}(a^{2}-b^{2})} - \frac{1}{2b^{2}(a^{2}-b^{2})} \cdot \frac{1}{a/b} \left[\tan^{-1}\frac{z}{a/b}\right]_{0}^{\infty} \)
\( = \frac{\pi}{4b^{2}(a^{2}-b^{2})} - \frac{\pi}{2ab(a^{2}-b^{2})} \)
\( = \frac{\pi}{4b^{2}(a^{2}-b^{2})} \left( \frac{1}{b}-\frac{1}{a} \right) \)
\( = \frac{\pi}{4ab^{2}(a+b)} \quad(\text{Proved}) \)
Solution:
\(r=2(1-\cos\theta)\)
\(\Rightarrow r^{2}=4(1+\cos^{2}\theta-2\cos\theta)\)
\(\frac{dr}{d\theta}=2\sin\theta\)
\(\left(\frac{dr}{d\theta}\right)^{2}=4\sin^{2}\theta\)
Now, \( \sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}} \)
\( =\sqrt{4(1+\cos^{2}\theta-2\cos\theta)+4\sin^{2}\theta} \)
\( =\sqrt{4+4(\sin^{2}\theta+\cos^{2}\theta)-8\cos\theta} \)
\( =\sqrt{4+4-8\cos\theta} \)
\( =\sqrt{8-8\cos\theta} \)
\( =2\sqrt{2}\sqrt{1-\cos\theta} \)
\( =2\sqrt{2}\sqrt{2\sin^{2}\frac{\theta}{2}} \)
\( =4\sin\frac{\theta}{2} \)
Now, \( I=\int_{0}^{\pi} \sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}} \,d\theta \)
\( =\int_{0}^{\pi} 4\sin\frac{\theta}{2}\,d\theta \)
\( =4\left[ \frac{-\cos\frac{\theta}{2}}{\frac{1}{2}} \right]_{0}^{\pi} \)
\( =-8(\cos\frac{\pi}{2}-\cos0) \)
\( =-8(0-1) \)
\( =8\quad(\text{Proved}) \)
Solution:
As, \(\tan^{-1}x=\frac{\pi}{2}-\cot^{-1}x\)
So, \( I=\frac{\pi}{2}-\int_{0}^{1}\cot^{-1}(1-x+x^{2})\,dx \)
\( =\frac{\pi}{2}-\int_{0}^{1}\tan^{-1}\left(\frac{1}{1-x+x^{2}}\right)dx \)
\( =\frac{\pi}{2}-\int_{0}^{1} \tan^{-1}\left( \frac{x-(x-1)}{1+x(x-1)} \right)dx \)
\( =\frac{\pi}{2}- \left[ \int_{0}^{1}\tan^{-1}x\,dx -\int_{0}^{1}\tan^{-1}(x-1)\,dx \right] \)
\( =\frac{\pi}{2} -\int_{0}^{1}\tan^{-1}x\,dx +\int_{0}^{1}\tan^{-1}(x-1)\,dx \)
As \(\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\)
\( \int_{0}^{1}\tan^{-1}(x-1)\,dx =\int_{0}^{1}\tan^{-1}(1-(x-1))\,dx \)
\( =\int_{0}^{1}\tan^{-1}(-x)\,dx =-\int_{0}^{1}\tan^{-1}x\,dx \)
Now, \( I=\frac{\pi}{2} -2\int_{0}^{1}\tan^{-1}x\,dx \)
\( =\frac{\pi}{2} -2\left[ x\tan^{-1}x -\frac{1}{2}\ln(1+x^{2}) \right]_{0}^{1} \)
\( =\frac{\pi}{2} -2\left[ \frac{\pi}{4} -\frac{1}{2}\ln2 \right] \)
\( =\frac{\pi}{2} -\frac{\pi}{2} +\ln2 \)
\( =\ln2\quad(\text{Ans}) \)
Solution:
Let \(x=\tan\theta\)
\(dx=\sec^{2}\theta\,d\theta\)
When \(x=0, \theta=0\) and when \(x=1, \theta=\frac{\pi}{4}\)
\( I=\int_{0}^{\pi/4} \frac{\log(1+\tan\theta)\sec^{2}\theta\,d\theta} {\sec^{2}\theta} \)
\( =\int_{0}^{\pi/4} \log(1+\tan\theta)\,d\theta \)
As, \(\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\)
\( I=\int_{0}^{\pi/4} \log\left(1+\tan\left(\frac{\pi}{4}-\theta\right)\right)d\theta \)
\( =\int_{0}^{\pi/4} \log\left(1+\frac{1-\tan\theta}{1+\tan\theta}\right)d\theta \)
\( =\int_{0}^{\pi/4} \log\left(\frac{2}{1+\tan\theta}\right)d\theta \)
\( =\int_{0}^{\pi/4}\log2\,d\theta -\int_{0}^{\pi/4}\log(1+\tan\theta)\,d\theta \)
\( I=\frac{\pi}{4}\log2-I \)
\( 2I=\frac{\pi}{4}\log2 \)
\( I=\frac{\pi}{8}\log2\quad(\text{Ans}) \)
Solution:
As, \(\int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx\)
\( I=\int_{3}^{6} \frac{\sqrt{x}\,dx} {\sqrt{x}+\sqrt{9-x}} \)
\( =\int_{3}^{6} \frac{\sqrt{3+6-x}\,dx} {\sqrt{3+6-x}+\sqrt{9-3-6+x}} \)
\( =\int_{3}^{6} \frac{\sqrt{9-x}} {\sqrt{9-x}+\sqrt{x}}\,dx \)
So, \( 2I=\int_{3}^{6} \frac{\sqrt{x}+\sqrt{9-x}} {\sqrt{9-x}+\sqrt{x}}\,dx \)
\( 2I=\int_{3}^{6}dx \)
\( I=\frac{1}{2}\int_{3}^{6}dx \)
\( =\frac{1}{2}[x]_{3}^{6} \)
\( =\frac{1}{2}(6-3) \)
\( =\frac{3}{2}\quad(\text{Proved}) \)
Solution:
\(I=\displaystyle\int_{0}^{\pi} \frac{x\,dx} {1+\cos x\sin\alpha}\)
As, \(\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx\)
\( I=\int_{0}^{\pi} \frac{\pi-x} {1+\cos(\pi-x)\sin\alpha}\,dx \)
\( =\int_{0}^{\pi} \frac{\pi-x} {1+\cos x\sin\alpha}\,dx \)
\( I=\pi\int_{0}^{\pi} \frac{dx} {1+\cos x\sin\alpha} -\int_{0}^{\pi} \frac{x\,dx} {1+\cos x\sin\alpha} \)
\( 2I=\pi\int_{0}^{\pi} \frac{dx} {1+\cos x\sin\alpha} \)
\( I=\frac{\pi}{2} \int_{0}^{\pi} \frac{dx} {1+\cos x\sin\alpha} \)
\( =\frac{\pi}{2} \int_{0}^{\pi} \frac{2\tan\frac{x}{2}} {(1+\tan^{2}\frac{x}{2}) +2\tan\frac{x}{2}\cos\alpha} \,dx \,[\because cosx=\frac{2tan \frac{x}{2}}{1+tan^2 \frac{x}{2}}] \)
\( =\frac{\pi}{2} \int_{0}^{\pi} \frac{(1+\tan^{2}\frac{x}{2})\,dx} {(1+\tan^{2}\frac{x}{2}) +2\tan\frac{x}{2}\cos\alpha} \)
\( =\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec^{2}\frac{x}{2}\,dx} {\tan^{2}\frac{x}{2} +2\tan\frac{x}{2}\cos\alpha+1} \)
Let \(\tan\frac{x}{2}=z\)
\(\frac{1}{2}\sec^{2}\frac{x}{2}\,dx=dz\)
\(\sec^{2}\frac{x}{2}\,dx=2dz\)
When \(x=0, z=0\) and when \(x=\pi, z=\infty\)
\( I=\pi \int_{0}^{\infty} \frac{dz} {z^{2}+2z\cos\alpha+1} \)
\( =\pi \int_{0}^{\infty} \frac{dz} {(z+\cos\alpha)^{2}+\sin^{2}\alpha} \)
\( =\frac{\pi}{\sin\alpha} \left[ \tan^{-1} \frac{z+\cos\alpha} {\sin\alpha} \right]_{0}^{\infty} \)
\( =\frac{\pi}{\sin\alpha} \left( \frac{\pi}{2} -\tan^{-1}\frac{\cos\alpha}{\sin\alpha} \right) \)
\( =\frac{\pi}{\sin\alpha} \left( \frac{\pi}{2} -\left(\frac{\pi}{2}-\alpha\right) \right) \)
\( =\frac{\pi\alpha}{\sin\alpha} \quad(\text{Proved}) \)
Solution:
\( \frac{x}{(x-1)^{2}(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{C}{x+2} \)
\( x = A(x-1)(x+2) + B(x+2) + C(x-1)^{2} \)
Putting \(x=1\),
\( 1=3B \Rightarrow B=\frac{1}{3} \)
Putting \(x=-2\),
\( -2=9C \Rightarrow C=-\frac{2}{9} \)
Putting \(x=0\),
\( 0=-2A+2B+C \)
\( 0=-2A+\frac{2}{3}-\frac{2}{9} \)
\( 0=-2A+\frac{4}{9} \Rightarrow A=\frac{2}{9} \)
Hence,
\( \int \frac{x\,dx} {(x-1)^{2}(x+2)} = \int \left( \frac{2/9}{x-1} + \frac{1/3}{(x-1)^{2}} - \frac{2/9}{x+2} \right)dx \)
\( = \frac{2}{9}\ln|x-1| -\frac{1}{3(x-1)} -\frac{2}{9}\ln|x+2| +C \)
Solution:
Break the interval at \(x=-5,-2,0\)
For \(-5\le x\le -2\), \( |x|=-x,\; |x+2|=-(x+2),\; |x+5|=x+5 \)
\( f(x)=-x-(x+2)+(x+5) =-x+3 \)
For \(-2\le x\le 0\), \( |x|=-x,\; |x+2|=x+2,\; |x+5|=x+5 \)
\( f(x)=-x+(x+2)+(x+5) =x+7 \)
Hence,
\( I=\int_{-5}^{-2}(-x+3)\,dx + \int_{-2}^{0}(x+7)\,dx \)
\( =\left[-\frac{x^{2}}{2}+3x\right]_{-5}^{-2} + \left[\frac{x^{2}}{2}+7x\right]_{-2}^{0} \)
\( =\left(4-6\right) -\left(\frac{25}{2}-15\right) + \left(0\right) -\left(2-14\right) \)
\( =(-2)-\left(-\frac{5}{2}\right)+12 \)
\( =\frac{1}{2}+12 =\frac{25}{2} \)
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