∫01 sin⁡2tan−1⁡1+x1−xdx is equal to

Let , I=∫01 sin⁡2tan−1⁡1+x1−xdxput, x=cos⁡2θThen, sin⁡2tan−1⁡1+cos⁡2θ1−cos⁡2θ=sin⁡2tan−1⁡(cot⁡θ)=sin⁡2tan−1⁡tan⁡π2−θ=sin⁡2π2−θ=sin⁡(π−2θ)=sin⁡2θ=1−cos2⁡2θ=1−x2Now,∫01 sin⁡(2tan−1⁡1+x1−xdx=∫01 1−x2dx=12x⋅1−x201+12sin−1⁡x01=12[11−1−0]+12sin−1⁡(1)−0=12[0]+12π2=π4

∫01 sin⁡2tan−1⁡1+x1−xdx is equal to

Let , I=∫01 sin⁡2tan−1⁡1+x1−xdxput, x=cos⁡2θThen, sin⁡2tan−1⁡1+cos⁡2θ1−cos⁡2θ=sin⁡2tan−1⁡(cot⁡θ)=sin⁡2tan−1⁡tan⁡π2−θ=sin⁡2π2−θ=sin⁡(π−2θ)=sin⁡2θ=1−cos2⁡2θ=1−x2Now,∫01 sin⁡(2tan−1⁡1+x1−xdx=∫01 1−x2dx=12x⋅1−x201+12sin−1⁡x01=12[11−1−0]+12sin−1⁡(1)−0=12[0]+12π2=π4

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$\int_{0}^{1} \sin (2 \tan^{- 1} \sqrt{\frac{1 + x}{1 - x}}) dx$ is equal to

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$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{2}$

answer is B.

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Detailed Solution

$\begin{matrix} Let, I = \int_{0}^{1} \sin (2 \tan^{- 1} \sqrt{\frac{1 + x}{1 - x}}) dx \\ put, x = \cos 2 θ \\ Then, \begin{aligned} \sin [2 \tan^{- 1} \sqrt{\frac{1 + \cos 2 θ}{1 - \cos 2 θ}}] = \sin [2 \tan^{- 1} (\cot θ)] \end{aligned} \\ = \sin [2 \tan^{- 1} \{\tan (\frac{π}{2} - θ)\}] \\ = \sin [2 (\frac{π}{2} - θ)] = \sin (π - 2 θ) = \sin 2 θ \\ = \sqrt{1 - \cos^{2} 2 θ} = \sqrt{1 - x^{2}} \end{matrix}$

Now,

$\begin{array}{l} \int_{0}^{1} \sin (2 \tan^{- 1} \sqrt{\frac{1 + x}{1 - x}}) dx = \int_{0}^{1} \sqrt{1 - x^{2}} dx \\ = {[\frac{1}{2} x \cdot \sqrt{1 - x^{2}}]}_{0}^{1} + \frac{1}{2} {[\sin^{- 1} x]}_{0}^{1} \\ = \frac{1}{2} [1 \sqrt{1 - 1} - 0] + \frac{1}{2} [\sin^{- 1} (1) - 0] \\ = \frac{1}{2} [0] + \frac{1}{2} (\frac{π}{2}) = \frac{π}{4} \end{array}$