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The value of ππcos2x1+axdx,a>0 is

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a
π/2
b
c
π
d

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detailed solution

Correct option is A

Let I=∫−ππ cos2⁡x1+axdxPut x=−θ to obtainI=∫π−π cos2⁡θ1+a−θ(−1)dθ=∫−ππ axcos2⁡x1+axdxAdding (1) and (2), we get 2I=∫−ππ cos2⁡xdx=2∫0π cos2⁡xdx=π⇒I=π/2


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