InterviewSolution
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`int_(0)^((pi)/(2))(sinx)/(1+cos^(2)x)dx` |
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Answer» माना `" "I=int_(0)^(pi//2)(sinx)/(1+cos^(2)x)dx` माना `" "cos x=t` `rArr -sin x dx=dt rArr sin x dx=- dt` `x =0" "rArr" "t=cos 0=1` और `" "x=(pi)/(2)" "rArr" "t=cos.(pi)/(2)=0` `therefore" "I=int_(1)^(0)(1)/(1+t^(2))(-dt)=-int_(1)^(0)(1)/(1+t^(2))dt` `=-[tan^(-1)t]_(1)^(0)` `=-[tan^(-1)(0)-tan^(-1)(1)]` `=-(0-(pi)/(4))=(pi)/(4)` |
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