Question

integrate sin^-1[2x/1+x^2]

Answer 1

I=∫arcsin(2x1+x2)dx.I=∫arcsin⁡(2x1+x2)dx.

Let x=tanθ⇒dx=sec2θdθ.x=tan⁡θ⇒dx=sec2⁡θdθ.

⇒2x1+x2=2tanθ1+tan2θ=2tanθsec2θ⇒2x1+x2=2tan⁡θ1+tan2⁡θ=2tan⁡θsec2⁡θ

=2sinθcosθ=sin2θ.=2sin⁡θcos⁡θ=sin⁡2θ.

⇒arcsin(2x1+x2)=arcsin(sin2θ)=2θ.⇒arcsin⁡(2x1+x2)=arcsin⁡(sin⁡2θ)=2θ.

⇒I=∫arcsin(2x1+x2)dx=∫2θsec2θdθ⇒I=∫arcsin⁡(2x1+x2)dx=∫2θsec2⁡θdθ

=2θtanθ−∫2tanθdθ=2θtanθ−2logsecθ+C=2θtan⁡θ−∫2tan⁡θdθ=2θtan⁡θ−2log⁡sec⁡θ+C

=2θtanθ−logsec2θ+C=2θtanθ−log(1+tan2θ)+C=2θtan⁡θ−log⁡sec2⁡θ+C=2θtan⁡θ−log⁡(1+tan2⁡θ)+C

=2xarctanx−log(1+x2)+C.

Answer 2

int(sin^-1x)^2dx=x(sin^-1x)^ 2+2sqrt(1-x^2)sin^-1x-2x+C We have: I=int( sin^-1x)^2dx Integration by parts takes the f