The integral \int_1 / 4^3 / 4 \cos (2 \cot ^-1 \sqrt\frac1-x1+x) d x is equal to

Definite Integration PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q36 with Solution

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Correct answer: B

Let $\theta=\cot^{-1}\sqrt{\frac{1-x}{1+x}}.$ Then the integrand is $\cos(2\theta).$
Use the identity $\cos 2\theta=\frac{\cot^2\theta-1}{\cot^2\theta+1}.$ Here, $\cot^2\theta=\frac{1-x}{1+x}.$ So, $\cos(2\theta)=\frac{\frac{1-x}{1+x}-1}{\frac{1-x}{1+x}+1}.$
Simplify: $\frac{\frac{1-x-(1+x)}{1+x}}{\frac{1-x+(1+x)}{1+x}}=\frac{\frac{-2x}{1+x}}{\frac{2}{1+x}}=-x.$ Hence the integral becomes $\int_{1/4}^{3/4} -x\,dx.$
Evaluate: $\int_{1/4}^{3/4} -x\,dx=-\left[\frac{x^2}{2}\right]_{1/4}^{3/4}.$ Now,

=-\frac{1}{2}\left(\frac{9}{16}-\frac{1}{16}\right) =-\frac{1}{2}\cdot\frac{8}{16} =-\frac{1}{4}.$$ 5. Therefore, $$\int_{1/4}^{3/4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) dx=-\frac{1}{4}.$$ 6. Option check: - A: $-\frac12$ ❌ - B: $-\frac14$ ✅ - C: $\frac14$ ❌ - D: $\frac12$ ❌ So the correct option is **B**.