The integral 1 4 3 4 cos ( 2 cot - 1 1 - x 1 + x ) d x is equal to [2024]

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Solution

Q.
$The integral \int_{\frac{1}{4}}^{\frac{3}{4}} \cos (2 \cot^{- 1} \sqrt{\frac{1 - x}{1 + x}}) d x is equal to$ [2024]

1 $\frac{1}{4}$

2 $- \frac{1}{4}$

3 $- \frac{1}{2}$

4 $\frac{1}{2}$

Ans.
(2)

$I = \int_{1 / 4}^{3 / 4} \cos (2 \cot^{- 1} \sqrt{\frac{1 - x}{1 + x}}) d x$

$= \int_{1 / 4}^{3 / 4} \cos (2 \tan^{- 1} \sqrt{\frac{1 + x}{1 - x}}) d x$

$= \int_{1 / 4}^{3 / 4} \frac{1 - \tan^{2} (\tan^{- 1} \sqrt{\frac{1 + x}{1 - x}})}{1 + \tan^{2} (\tan^{- 1} \sqrt{\frac{1 + x}{1 - x}})} d x$

$= \int_{1 / 4}^{3 / 4} \frac{1 - \frac{1 + x}{1 - x}}{1 + \frac{1 + x}{1 - x}} d x$

$= \int_{1 / 4}^{3 / 4} \frac{- 2 x}{2} d x = {(- \frac{x^{2}}{2})}_{1 / 4}^{3 / 4} = \frac{- 9}{32} + \frac{1}{32} = \frac{- 8}{32} = \frac{- 1}{4}$

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