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Question: For any integer \(\int _ { 0 } ^ { \pi } e ^ { \sin ^ { 2 } x } \cos ^ { 3 } ( 2 n + 1 ) x d x =\)...

For any integer ∫0πesin⁡2xcos⁡3(2n+1)xdx=\int _ { 0 } ^ { \pi } e ^ { \sin ^ { 2 } x } \cos ^ { 3 } ( 2 n + 1 ) x d x =

A

−1- 1

B

0

C

1

D

π\pi

Answer

0

Explanation

Solution

Let f(x)=∫0πesin⁡2xcos⁡3(2n+1)x⋅dxf ( x ) = \int _ { 0 } ^ { \pi } e ^ { \sin ^ { 2 } x } \cos ^ { 3 } ( 2 n + 1 ) x \cdot d x

Since cos⁡(2n+1)(π−x)=cos⁡[(2n+1)π−(2n+1)x]\cos ( 2 n + 1 ) ( \pi - x ) = \cos [ ( 2 n + 1 ) \pi - ( 2 n + 1 ) x ]

=−cos⁡(2n+1)x= - \cos ( 2 n + 1 ) xand sin⁡2(π−x)=sin⁡2x\sin ^ { 2 } ( \pi - x ) = \sin ^ { 2 } x

Hence by the property of definite integral,

∫0πesin⁡2xcos⁡3(2n+1)xdx=0\int _ { 0 } ^ { \pi } e ^ { \sin ^ { 2 } x } \cos ^ { 3 } ( 2 n + 1 ) x d x = 0, [f(2a−x)=−f(x)][ f ( 2 a - x ) = - f ( x ) ] .