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int(0)^(pi//2)sinx sin2x sin3x sin 4x dx...

`int_(0)^(pi//2)sinx sin2x sin3x sin 4x dx=`

A

`(pi)/(4)`

B

`(pi)/(8)`

C

`(pi)/(16)`

D

`(pi)/(32)`

Text Solution

Verified by Experts

The correct Answer is:
C
`I=int_(0)^(pi//2)sinx sin2x sin3x sin 4x dx=int_(0)^(pi//2)cos x sin 2x cos3x sin 4x dx`
`rArr" "2I=int_(0)^(pi//2)cos2xsin2xsin4xdx`
`=(1)/(2)int_(0)^(pi//2)sin^(2)4xdx`
`=(1)/(2)int_(0)^(pi//2)(1-cos8x)/(2)dx`
`=(1)/(4)[x-(sin8x)/(8)]_(0)^(pi//2)=(pi)/(8)`
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