How do you find the indefinite integral of #int cosx/sqrtsinx#?

Question

4180 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-11T16:17:28

Let #u = sinx#.

#du = cosx(dx)#

#dx = (du)/cosx#

#=int(cosx/u^(1/2))dx#

#=int(cosx/u^(1/2))((du)/cosx)#

#=int(u^(-1/2))du#

#=2u^(1/2) + C#

#=2(sinx)^(1/2) + C#

#=2sqrt(sinx) + C#

You can check your result by differentiating. You will get the initial function

Hopefully this helps!