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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Inverse Trigonometric Functions

The integral ∫_ 1/4 ^ 3 / 4 (2 ^ -1 (1-x)/(1+x) ) d x is equal to

The integral ∫1/43/4cos⁡(2cot⁡−11−x1+x)dx\int_{1/4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x is equal to

Options

  1. A1/21 / 2
  2. B−1/2-1 / 2
  3. C−1/4-1 / 4
  4. D1/41 / 4

Correct answer

C. −1/4-1 / 4

Step-by-step solution

I=∫1/43/4cos⁡(2cot⁡−1(1−x1+x)dx)∫1/43/4cos⁡(2(tan⁡−11+x1+x))dx∫1/43/41−tan⁡2(tan⁡−11+x1−x)1+tan⁡2(tan⁡−11+x1−x)dx\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}=∫1/43/41−(1+x1−x)1+(1+x1−x)dx=∫1/43/4−2x2dx=∫1/43/4(−x)dx=−(x22)1/43/4=−12[916−116]=−14\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}$
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