InterviewSolution
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InterviewSolution
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Evaluate `int_(0)^(pi)(x)/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx`. |
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Answer» Let `I=int_(0)^(pi)(x)/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx`…….`(i)` Then, `I=int_(0)^(pi)((pi-x))/((a^(2)cos^(2)(pi-x)+b^(2)sin^(2)(pi-x)])dx` or `I=int_(0)^(pi)((pi-x))/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx`........`(ii)` Adding `(i)` and `(ii)`, we get `2I=int_(0)^(pi)((x+pi-x))/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx=piint_(0)^(pi)(dx)/((a^(2)cos^(2)x+b^(2)sin^(2)x))` `=2piint_(0)^(pi//2)(dx)/((a^(2)cos^(2)x+b^(2)sin^(2)x))=2piint_(0)^(pi//2)(sec^(2)x)/((a^(2)+b^(2)tan^(2)x))dx` [dividing num. and denom. by `cos^(2)x`] `=2piint_(0)^(oo)(dt)/((a^(2)+b^(2)t^(2)))`, where `tanx=t` `=(2pi)/(b^(2))int_(0)^(oo)(dt)/(((a^(2))/(b^(2))+t^(2)))=[(2pi)/(b^(2))*(b)/(a)tan^(-1)((bt)/(a))]_(0)^(oo)` `=(2pi)/(ab)[tan^(-1)(oo)-tan^(-1)(0)]=(2pi)/(ab)((pi)/(2)-0)=((2pi)/(ab)xx(pi)/(2))=(pi^(2))/(ab)`. `:.I=(pi^(2))/(2ab)impliesint_(0)^(pi)(x)/((a^(2)cos^(2)x+b^(2)sin^(2)x))dx=(pi^(2))/(2ab)`. |
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