`int_(0)^(1)(x tan^(-1)x)/((1+x^(2))^(3//2))dx` का मान ज्ञात कीजिए ।

Explanation

माना `x=tan theta`
`rArr" "dx=sec^(2) theta d theta`
`{:(x=0," पर ",tan theta=0,rArr.,theta=0),(x=1," पर ",tan theta=1,rArr.,theta=pi//4):}`
`therefore int_(0)^(1)(x tan^(-1)x)/((1+x^(2))^(3//2))dx`
`=int_(0)^(pi//4)(theta tan theta)/((1+ tan^(2) theta)^(3//2)).sec^(2) theta d theta`
`=int_(0)^(pi//4)theta sin theta d theta`
`=[- theta cos theta]_(0)^(pi//4)+int_(0)^(pi//4)1. cos theta d theta`
`=-((pi)/(4) cos.(pi)/(4)-0)+[sin theta]_(0)^(pi//4)`
`=-(pi)/(4sqrt2)+(1)/(sqrt2)=(sqrt2(4-pi))/(8)`