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Mathematics · Definite Integration

JEE Main 2024 — 9 April, Shift 2 — Question 15

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:

  1. Option A:

    −1/2-1 / 2

  2. Option B:

    1/41 / 4

  3. Option C:

    1/21 / 2

  4. Option D:

    −1/4-1 / 4

    Correct

Answer: D

Step-by-step solution

I=∫1/43/4cos⁡(2cot⁡−1(1−x1+x)dx)\quad I=\int_{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1}\left(\sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}}\right) \mathrm{dx}\right)

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

∫1/43/41−tan⁡2(tan⁡−11+x1−x)1+tan⁡2(tan⁡−11+x1−xdx\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.} d x

=∫1/43/41−(1+x1−x)1+(1+x1−x)dx=∫1/43/4−2x2dx=\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 =∫1/43/4(−x)dx=−(x22)1/43/4=\int_{1 / 4}^{3 / 4}(-x) d x=-\left(\frac{x^{2}}{2}\right)_{1 / 4}^{3 / 4}

=−12[916−116]=-\frac{1}{2}\left[\frac{9}{16}-\frac{1}{16}\right] =−14=-\frac{1}{4}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Mathematics
Chapter
Definite Integration
Topic
Evaluation of Definite Integrals

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