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Question

# ∫3π−3πsin2θ.sin22θ dθ is

A
π
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B
3π2
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C
3π2
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D
6π
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Solution

## The correct option is C 3π2I=3π∫−3πsin2θ.sin22θ dθ We know that if f is an even function ⇒a∫−af(x)dx=2a∫0f(x)dx I=23π∫0(sinθ.sin2θ)2 dθ I=123π∫0(2sinθ.sin2θ)2 dθ we know that 2sinA.sinB=cos(A−B)−cos(A+B) I=123π∫0(cosθ−cos3θ)2 dθ I=123π∫0(cos2θ−2cos3θcosθ+cos23θ) dθ Using the formula2cosA.cosB=cos(A+B)+cos(A−B), we get I=123π∫0(1+cos2θ2−(cos4θ+cos2θ)+1+cos6θ2) dθ I=143π∫0(2+cos6θ−2cos4θ−cos2θ) dθ I=14[3π∫02 dθ+3π∫0cos6θ dθ−3π∫02cos4θ dθ−3π∫0cos2θ dθ] I=14[2×(3π−0)+(0−0)−2×(0−0)−(0−0)] I=6π4=3π2

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