Evaluate the following integrals: 2 ^ -1 1+x 1-x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$

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Answer

Let $\text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}$
Let $\text{x}=\cos2\theta$
On differentiating both sides, we get
$\text{dx}=-2\sin2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\cos2\theta}{1-\cos2\theta}}\Big\}2\sin2\theta\text{ d}\theta$
$=-2\int\cos\Big\{2\cot^{-1}\sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}}\Big\}\sin2\theta\text{ d}\theta$
$=-2\int\cos\{2\cot^{-1}(\cot\theta)\}\sin2\theta\text{ d}\theta$
$=-2\int\cos2\theta\sin2\theta\text{ d}\theta$
$=\frac{\cos4\theta}{4}+\text{C}_1$
$=-\int\sin4\theta\text{ d}\theta$
$=\frac{1}{4}(2\cos^2\theta-1)+\text{C}_1$
$=\frac{1}{2}\text{x}^2-\frac{1}{4}+\text{C}_1$
$=\frac{1}{2}\text{x}^2+\text{C},$ where $\text{C}=-\frac{1}{4}+\text{C}_1$
Hence, $\int\cos\Big\{2\cot^{-1}\sqrt{\frac{1+\text{x}}{1-\text{x}}}\Big\}\text{dx}=\frac{1}{2}\text{x}^2+\text{C}$

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