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प्रश्न
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = ______.
विकल्प
`pi/2 + log 2`
`pi/2 - log 2`
`pi/2 + 1/2 log 2`
`pi/2 - 1/2 log 2`
MCQ
रिक्त स्थान भरें
उत्तर
`int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x` = `pi/2 - log 2`.
Explanation:
Let I = `int_0^1 sin^-1 ((2x)/(1 + x^2))"d"x`
Put x = θ
⇒ dx = sec2θ dθ
∴ I = `int_0^(pi/4) sin^1 ((2tantheta)/(1 ++ tan^2theta))sec^2theta "d"theta`
= `int_0^(pi/4) sin^-1(sin 2theta) sec^2theta "d"theta`
= `int_0^(pi/4) 2theta sec^3theta "d"theta`
= `2 int_0^(pi/4) theta sec^2 theta "d"theta`
= `2[[theta tan theta]_0^(pi/4) - int_0^(pi/4) 1*tan theta "d'theta]`
= `2[(pi/4 tan pi/4 - 0) - [log|sectheta|]_0^(pi/4)]`
= `2[pi/4 - (log|sec pi/4| - log|sec0|)]`
= `2[pi/4 - (log sqrt(2) - log 1)]`
= `2(pi/4 - 1/2 log 2)`
∴ I = `pi/2 - log 2`
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