Q44 / 90 · JEE Main 2024 (09 Apr Shift 2) · Mathematics

The integral $\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x$ is equal to

  • A
    $1 / 2$
  • B
    $-1 / 2$
  • Correct answer
    C
    $-1 / 4$
  • D
    $1 / 4$

Solution

$\begin{aligned} & I=\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) d x\right) \\ & \int_{1 / 4}^{3 / 4} \cos \left(2\left(\tan ^{-1} \sqrt{\frac{1+x}{1+x}}\right)\right) d x \\ & \left.\int_{1 / 4}^{3 / 4} \frac{1-\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right)}{1+\tan ^2\left(\tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right.}\right) d x\end{aligned}$ $\begin{aligned} & =\int_{1 / 4}^{3 / 4} \frac{1-\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)} d x=\int_{1 / 4}^{3 / 4} \frac{-2 x}{2} d x \\ & =\int_{1 / 4}^{3 / 4}(-x) d x=-\left(\frac{x^2}{2}\right)_{1 / 4}^{3 / 4} \\ & =-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right] \\ & =-\frac{1}{4}\end{aligned}$

Hence, the correct answer is option: 'C'.

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