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Question: \(\int _ { 0 } ^ { 1 } \frac { d x } { x ^ { 2 } + 2 x \cos \alpha + 1 }\) =...

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

A

sina

B

a sin a

C

α2sin⁡α\frac { \alpha } { 2 \sin \alpha }

D

α2\frac { \alpha } { 2 }sin a

Answer

α2sin⁡α\frac { \alpha } { 2 \sin \alpha }

Explanation

Solution

I =

=1sin⁡α= \frac { 1 } { \sin \alpha } [tan⁡−1(x+cos⁡αsin⁡α)]01\left[ \tan ^ { - 1 } \left( \frac { \mathrm { x } + \cos \alpha } { \sin \alpha } \right) \right] _ { 0 } ^ { 1 }

=1sin⁡α[tan⁡−1(1+cos⁡αsin⁡α)−tan⁡−1(cos⁡αsin⁡α)]\frac { 1 } { \sin \alpha } \left[ \tan ^ { - 1 } \left( \frac { 1 + \cos \alpha } { \sin \alpha } \right) - \tan ^ { - 1 } \left( \frac { \cos \alpha } { \sin \alpha } \right) \right]

= 1sin⁡α[π2−cot⁡−1(cot⁡α2)−(π2−cot⁡−1(cot⁡α))]\frac { 1 } { \sin \alpha } \left[ \frac { \pi } { 2 } - \cot ^ { - 1 } \left( \cot \frac { \alpha } { 2 } \right) - \left( \frac { \pi } { 2 } - \cot ^ { - 1 } ( \cot \alpha ) \right) \right]

= 1sin⁡α[−α2+α]=α2sin⁡α\frac { 1 } { \sin \alpha } \left[ - \frac { \alpha } { 2 } + \alpha \right] = \frac { \alpha } { 2 \sin \alpha }