How do you find#intarcsinx/sqrt(1-x^2)dx#?

Question

1717 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-22T14:09:28

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

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

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

#=tdt#,

#=1/2t^2#,

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

Answer 2 · 2018-07-22T14:11:42

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

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#

Answer 3 · 2018-07-22T14:12:13

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

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#