InterviewSolution
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InterviewSolution
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`int_0^pixsin^3x dx` |
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Answer» Let `I=int_(0)^(pi) x sin^(3)xdx`…….`(i)` Then, `I=int_(0)^(pi)(pi-x)sin^(3)(pi-x)dx` or `I=int_(0)^(pi)(pi-x)sin^(3)xdx`……..`(ii)` Adding `(i)` and `(ii)` , we get `2I=int_(0)^(pi)pisin^(3)xdx=piint_(0)^(pi)sin^(2)x*sinxdx` `=pi int_(0)^(pi)(1-cos^(2)x)sinxdx` `=-pi int_(1)^(-1)(1-t^(2))dt`, where `cosx=t` [`x=0impliest=1` and `x=piimpliest=-1`] `=piint_(-1)^(1)(1-t^(2))dt=pi[t-(t^(3))/(3)]_(-1)^(1)=(4pi)/(3)`. Hence, `I=(2pi)/(3)impliesint_(0)^(pi)xsin^(3)xdx=(4pi)/(3)`. |
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