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Question

# The value of π∫−π2x(1+sin x)1+cos2x dx is

A
π24
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B
π2
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C
0
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D
π2
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Solution

## The correct option is B π2I=π∫−π2x(1+sin x)1+cos2x dx =π∫−π2x1+cos2xdx+2π∫−πxsin x1+cos2xdx⋯(i) We know that, a∫−af(x) dx=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩2a∫0f(x) dx, if f(x) is even0, if f(x) is odd =0+4π∫0xsinx1+cos2xdx =4π∫0(π−x)sin(π−x)1+cos2(π−x) =4π∫0(π−x)sinx1+cos2xdx =4ππ∫0sinx1+cos2xdx−4π∫0xsinx1+cos2xdx ⇒2I=4ππ∫0sinx1+cos2xdx ⇒I=2ππ∫0sinx1+cos2xdx put cosx=t⇒−sinx dx=dt When x=0,t=1; and x=π,t=−1. ⇒I=2π−1∫1−dt1+t2 ⇒I=2π1∫−1dt1+t2 =4π[tan−1t]10 =4π⋅π4=π2

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