Our school exam need substitution methodlike, x = sin1/2θ, cosΘ

Dear aby

you can also solve by subtituting x²=sin2Θ

tan^-1[(1+x2)^1/2 - (1-x2)^1/2]/[(1+x2)^1/2 + (1-x2)^1/2]

=tan^-1[(1+sin2Θ)^1/2 - (1-sin2Θ)^1/2]/[(1+sin2Θ)^1/2 + (1-sin2Θ)^1/2]

now 1+sin2Θ =sin²Θ +cos²Θ + 2sinΘ cosΘ

          1+sin2Θ    =(sinΘ + cosΘ)²

      similerly 1-sin2Θ =(sinΘ - cosΘ)²

^{put these value}

tan^-1[(sinΘ+cosΘ) - (sinΘ -cosΘ)]/[(sinΘ+cosΘ) + (sinΘ -cosΘ)]

or    =   tan^-1 cotΘ

       = ∏/2 - Θ

       =∏/2 - 1/2 sin^-1 x²

now differentiate and you will get the result

Please feel free to post as many doubts on our discussion forum as you can.
 If you find any question Difficult to understand - post it here and we will get you the

answer and detailed solution very quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin