Question

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

Answer 1

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!