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Question

Question: \(\int _ { 0 } ^ { \pi / 2 } \frac { \sin x \cos x } { 1 + \sin ^ { 4 } x } d x =\)...

∫0π/2sin⁡xcos⁡x1+sin⁡4xdx=\int _ { 0 } ^ { \pi / 2 } \frac { \sin x \cos x } { 1 + \sin ^ { 4 } x } d x =

A

π2\frac { \pi } { 2 }

B

π4\frac { \pi } { 4 }

C

π6\frac { \pi } { 6 }

D

π8\frac { \pi } { 8 }

Answer

π8\frac { \pi } { 8 }

Explanation

Solution

Put sin⁡2x=t⇒dt=2sin⁡xcos⁡xdx\sin ^ { 2 } x = t \Rightarrow d t = 2 \sin x \cos x d x

Now ∫0π/2sin⁡xcos⁡x1+sin⁡4xdx=12∫0111+t2dt=12[tan⁡−1t]01=π8\int _ { 0 } ^ { \pi / 2 } \frac { \sin x \cos x } { 1 + \sin ^ { 4 } x } d x = \frac { 1 } { 2 } \int _ { 0 } ^ { 1 } \frac { 1 } { 1 + t ^ { 2 } } d t = \frac { 1 } { 2 } \left[ \tan ^ { - 1 } t \right] _ { 0 } ^ { 1 } = \frac { \pi } { 8 } .