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Question: \(\int _ { 0 } ^ { 1 } \sin \left( 2 \tan ^ { - 1 } \sqrt { \frac { 1 + x } { 1 - x } } \right) d x ...

∫01sin⁡(2tan⁡−11+x1−x)dx=\int _ { 0 } ^ { 1 } \sin \left( 2 \tan ^ { - 1 } \sqrt { \frac { 1 + x } { 1 - x } } \right) d x =

A

π/6\pi / 6

B

π/4\pi / 4

C

π/2\pi / 2

D

π\pi

Answer

π/4\pi / 4

Explanation

Solution

∫01sin⁡(2tan⁡−11+x1−x)dx\int _ { 0 } ^ { 1 } \sin \left( 2 \tan ^ { - 1 } \sqrt { \frac { 1 + x } { 1 - x } } \right) d x

Put sin⁡[2tan⁡−11+cos⁡θ1−cos⁡θ]\sin \left[ 2 \tan ^ { - 1 } \sqrt { \frac { 1 + \cos \theta } { 1 - \cos \theta } } \right]

=sin⁡[2tan⁡−1(cot⁡θ2)]= \sin \left[ 2 \tan ^ { - 1 } \left( \cot \frac { \theta } { 2 } \right) \right]

=sin⁡[2tan⁡−1[tan⁡(π2−θ2)]]=sin⁡[2(π2−θ2)]= \sin \left[ 2 \tan ^ { - 1 } \left[ \tan \left( \frac { \pi } { 2 } - \frac { \theta } { 2 } \right) \right] \right] = \sin \left[ 2 \left( \frac { \pi } { 2 } - \frac { \theta } { 2 } \right) \right]

=sin⁡(π−θ)=sin⁡θ=1−cos⁡2θ=1−x2= \sin ( \pi - \theta ) = \sin \theta = \sqrt { 1 - \cos ^ { 2 } \theta } = \sqrt { 1 - x ^ { 2 } }

Now, ∫01sin⁡(2tan⁡−11+x1−x)dx=∫011−x2dx\int _ { 0 } ^ { 1 } \sin \left( 2 \tan ^ { - 1 } \sqrt { \frac { 1 + x } { 1 - x } } \right) d x = \int _ { 0 } ^ { 1 } \sqrt { 1 - x ^ { 2 } } d x