How do you integrate #int cos x / ((sin x)^(1/2))dx#?

Question

3098 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-02T23:34:55

#int cosx/(sinx)^(1/2) dx=2(sinx)^(1/2)+C#

Let's try to guess what was derived.

We could notice, that:
#d/dx sinx=cosx# and #d/dx (f(x)^(1/2))=(d/dx f(x))/(2f(x)^(1/2))# (chain rule)

So we try to substitute #u=sinx#:
#du=cosx dx#

#int cosx/(sinx)^(1/2) dx=int (du)/u^(1/2)=2int (du)/(2u^(1/2))=2u^(1/2)+C#

We need to substitute #u=sinx# back:

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

Please tell if you need more explanation.