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Question: The value of the definite integral \(\int _ { 0 } ^ { 1 } \frac { d x } { x ^ { 2 } + 2 x \cos \alp...

The value of the definite integral ∫01dxx2+2xcos⁡α+1\int _ { 0 } ^ { 1 } \frac { d x } { x ^ { 2 } + 2 x \cos \alpha + 1 } for 0<α<π0 < \alpha < \pi is equal to

A

sin⁡α\sin \alpha

B

tan⁡−1(sin⁡α)\tan ^ { - 1 } ( \sin \alpha )

C

αsin⁡α\alpha \sin \alpha

D

α2(sin⁡α)−1\frac { \alpha } { 2 } ( \sin \alpha ) ^ { - 1 }

Answer

α2(sin⁡α)−1\frac { \alpha } { 2 } ( \sin \alpha ) ^ { - 1 }

Explanation

Solution

∫01dxx2+2xcos⁡α+1=∫01dx(x+cos⁡α)2+1−cos⁡2α\int _ { 0 } ^ { 1 } \frac { d x } { x ^ { 2 } + 2 x \cos \alpha + 1 } = \int _ { 0 } ^ { 1 } \frac { d x } { ( x + \cos \alpha ) ^ { 2 } + 1 - \cos ^ { 2 } \alpha }

=∫01dx(x+cos⁡α)2+sin⁡2α=[1sin⁡αtan⁡−1x+cos⁡αsin⁡α]01= \int _ { 0 } ^ { 1 } \frac { d x } { ( x + \cos \alpha ) ^ { 2 } + \sin ^ { 2 } \alpha } = \left[ \frac { 1 } { \sin \alpha } \tan ^ { - 1 } \frac { x + \cos \alpha } { \sin \alpha } \right] _ { 0 } ^ { 1 }

=1sin⁡α(tan⁡−1cot⁡α2−tan⁡−1cot⁡α)=α2⋅1sin⁡α= \frac { 1 } { \sin \alpha } \left( \tan ^ { - 1 } \cot \frac { \alpha } { 2 } - \tan ^ { - 1 } \cot \alpha \right) = \frac { \alpha } { 2 } \cdot \frac { 1 } { \sin \alpha } .