InterviewSolution
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InterviewSolution
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Evaluate: `int_0^1(x t a n^(-1)x)/((1+x^2)^(3//2))dx` |
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Answer» `(i)` Put `x=tantheta` so that `dx=sec^(2)theta d theta`. Clearly, `x=0impliestheta=0` and `x=1impliestheta=(pi//4)`. `:.int_(0)^(1)(xtan^(-1)x)/((1+x^(2))^(3//2))dx=int_(0)^(pi//4)(theta tan theta)/(sec^(3)theta)*sec^(2)theta d theta=int_(0)^(pi//4)theta sin d theta` `=[-thetacostheta]_(0)^(pi//4)-int_(0)^(pi//4)(-costheta)d theta`[integrating by parts] `=[-thetacostheta]_(0)^(pi//4)+[sintheta]_(0)^(pi//4)=-(pi)/(4)cos.(pi)/(4)+sin.(pi)/(4)` `=((-pi)/(4sqrt(2))+(1)/(sqrt(2)))=(4-pi)/(4sqrt(2))=(sqrt(2)(4-pi))/(8)`. `(ii)` Put `x=sin theta` so that `dx=costheta d theta` Clearly, `(x=0impliestheta=0)` and `(x=(1)/(sqrt(2))impliestheta=(pi)/(4))`. `:.int_(0)^(1//sqrt(2))(sin^(-1)x)/((1-x^(2))^(3//2))dx=int_(0)^(pi//4)(theta)/(cos^(3)theta)*costhetad theta=int_(0)^(pi//4)theta sec^(2)theta d theta` `=[theta tan theta]_(0)^(pi//4)-int_(0)^(pi//4)1*tan theta d theta`[integrating by parts] `=(pi)/(4)+[log(costheta)]_(0)^(pi//4)=(pi)/(4)+log(cos.(pi)/(4))` `=(pi)/(4)+log((1)/(sqrt(2)))=((pi)/(4)-(1)/(2)log2)`. |
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