InterviewSolution
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InterviewSolution
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`int_0^pi(x dx)/(a^2cos^2x+b^2sin^2x)`A. `pi/(2ab)`B. `pi/(ab)`C. `(pi^(2))/(2ab)`D. `(pi^(2))/(ab)` |
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Answer» Correct Answer - C Let `l=int_(0)^(pi)(xdx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)" ...(i)"` `rArr" "l=int_(0)^(pi)((pi-x)dx)/(a^(2)cos^(2)x+b^(2)sin^(2)x)" ...(ii)"` On adding Eqs. (i) and (ii), we get `2l=2piint_(0)^(pi//2)(dx)/(a^(2)cos^(2)x+b^(2)sin 2x)` `rArr" "2l=2piint_(0)^(pi//2)(sec^(2)x)/(a^(2)+b^(2)tan^(@)x)` Now, put `tanx=t rArr dx=(dt)/(sec^(2)x)` `therefore" "2l=2piint_(0)^(oo)(dt)/(a^(2)+b^(2)t^(2))` `=(1)/(b^(2))xx2piint_(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)xx(pi)/(2)` `rArr" "l=(pi^(2))/(2ab)` |
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