InterviewSolution
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InterviewSolution
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Evaluate `int_(0)^(pi)(x)/((1+sinx))dx`. |
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Answer» Let `I=int_(0)^(pi)(x)/((1+sinx))dx`…..`(i)` Then, `I=int_(0)^(pi)((pi-x))/(1+xsin(pi-x))dx=int_(0)^(pi)((pi-x))/((1+sinx))dx`…….`(ii)` Adding `(i)` and `(ii)`, we get `2I=pi int_(0)^(pi)(dx)/((1+sinx))=pi*int_(0)^(pi)(1)/((1+sinx))xx((1-sinx))/((1-sinx))dx` or `2I=pi int_(0)^(pi)((1-sinx)/(cos^(2)x))dx=pi*[int_(0)^(pi)sec^(2)xdx-int_(0)^(pi)secxtanxdx]` `=pi*{[tanx]_(0)^(pi)-[secx]_(0)^(pi)}=2pi` `:.I=pi`, i.e, `int_(0)^(pi)(x)/((1+sinx))dx=pi`. |
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