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Question

The value of 2π0xsin8xsin8x+cos8xdx is equal to :

A
2π
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B
4π
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C
2π2
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D
π2
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Solution

The correct option is D π2
Let I=2π0xsin8xsin8x+cos8xdx ...(1)I=2π0(2πx)sin8(2πx)sin8(2πx)+cos8(2πx)dx=2π0(2πx)sin8xsin8x+cos8xdx ...(2)

Adding (1) & (2), we get :

2I=2π2π0sin8xsin8x+cos8xdxI=π2π0sin8xsin8x+cos8xdx
I=4ππ20sin8xsin8x+cos8xdx ...(3)
I=4ππ20sin8(π2x)sin8(π2x)+cos8(π2x)dxI=4ππ20cos8xsin8x+cos8xdx ...(4)
Adding (3) and (4) , we get :
I=2ππ201 dx=2π×π2=π2

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