How do you findintarcsinx/sqrt(1-x^2)dxarcsinx1x2dx?

3 Answers
Jul 22, 2018

1/2(arcsinx)^2+C12(arcsinx)2+C.

Explanation:

If we subst. arcsinx=t, 1/sqrt(1-x^2)dx=dtarcsinx=t,11x2dx=dt.

:. int(arcsinx)/sqrt(1-x^2)dx=intarcsinx*1/sqrt(1-x^2)dx,

=tdt,

=1/2t^2,

=1/2(arcsinx)^2+C.

Jul 22, 2018

The answer is =(arcsinx)^2/2+C

Explanation:

Perform this integral by substitution

Let u=arcsinx, =>, du=(dx)/sqrt(1-x^2)

The integral is

I=int(arcsinxdx)/sqrt(1-x^2)

=intudu

=u^2/2

=(arcsinx)^2/2+C

Jul 22, 2018

int(arc sinx)/sqrt(1-x^2)dx=(arcsinx)^2/2+c

Explanation:

Here,

I=int(arc sinx)/sqrt(1-x^2)dx=intarc sinx*1/sqrt(1-x^2)dx

Subst. color(blue)(arc sinx=u=>1/sqrt(1-x^2)dx=du

:.I=intudu

=>I=u^2/2+c

Subst. back color(blue)(u=arc sinx

:.I=(arcsinx)^2/2+c