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

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Inverse Trigonometric Functions

The integral

\int_{1/4}^{3/4} cos (2 cot^{- 1} \frac{1 - x}{1 + x}) dx

is equal to:

Answer: (D)

Let

θ = cot^{- 1} \frac{1 - x}{1 + x}

, so

cot^{2} θ = \frac{1 - x}{1 + x}

cos 2 θ = \frac{cot ^{2} θ - 1}{cot ^{2} θ + 1} = \frac{\frac{1 - x}{1 + x} - 1}{\frac{1 - x}{1 + x} + 1} = \frac{\frac{- 2 x}{1 + x}}{\frac{2}{1 + x}} = - x

I = \int_{1/4}^{3/4} (- x) d x = - [\frac{x ^{2}}{2}]_{1/4}^{3/4} = - \frac{1}{2} (\frac{9}{16} - \frac{1}{16}) = - \frac{1}{2} \cdot \frac{1}{2} = - \frac{1}{4}

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

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