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`f(x)=int_1^x(tan^(-1)(t))/t dtAAx in R^+,t h e nfin dt h ev a l u eof``f(e^2)-f(1/(e^2))` |
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Answer» Correct Answer - `pi` `f(x)=int_(1)^(x)(tan^(-1)(t))/t dt` `:.f(1/x)=int_(1)^(1//x)(tan^(-1)(t))/tdt` Put `=t=1//u` `:.dt=-(du)/(u^(2))` `:.f(1//x)=int_(1)^(x)("tan"^(-1)(1/u))/(1/u)(-1/(u^(2)))du` `=-int_(1)^(x)("tan"^(-1)(1/u))/u du` `=-int_(1)^(x)("cot"^(-1)(u))/u du` `=-int_(1)^(x)(cot^(-1)(t))/t dt` Now` f(x)-f(1//x)=int_(1)^(x)(tan^(-1)t+cot^(-1)t)/t dt` `=int_(1)^(x) (pi)/2xx1/t dt` `=(pi)/2 log (x)` `:. f(e^(2))-f(1//e^(2))=(pi)/2log_(e)e^(2)=pi` |
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