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For 0<α<π he value of the definite integral  01dx/x2+2xcosα+1 is equal to

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a
α/(2sin⁡α)
b
tan−1⁡(sin⁡α)
c
αsin⁡α
d
a/2αsin⁡α

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detailed solution

Correct option is A

∫01 dxx2+2xcos⁡α+1=∫01 dx(x+cos⁡α)2+1−cos2⁡α=∫01 dx(x+cos⁡α)2+sin2⁡α=1sin⁡αtan−1⁡x+cos⁡αsin⁡α01=1sin⁡αtan−1⁡1+cos⁡αsin⁡α−tan−1⁡cos⁡αsin⁡α=1sin⁡αtan−1⁡cot⁡α2−tan−1⁡cot⁡α=1sin⁡απ2−α2−π2+α=α2sinα


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