Integral of Logarithms + Trig Functions

[Blues_MTA]

Homework Statement

Sec(x)/((ln(tan(x)+sec(x))^1/2)

We were instructed to find the integral

Homework Equations

Here is a link to the wolfram solution, i don't understand the steps they tookhttp://www.wolframalpha.com/input/?i=Integral+of+Sec%28x%29%2F%28%28ln%28sec%28x%29%2Btan%28x%29%29^1%2F2%29%29

The Attempt at a Solution

I understand that using the substitution method using u = Ln(tan(x)+sec(x))

du = Sec(x)^2+tan(x)Sec(x)/Sec(x)+Tan(x) dx

I do not understand how to substitute this and get 1/((u)^1/2) which is according to wolfram, I don't understand how the substitution method eliminates the Sec from the numerator, Later they evaluate that 1/(u^(1/2)) as 2((u)^1/2) can someone please explain these intermediate steps?

 

[micromass]

Observe that du=sec(x)dx. Can you do it now?

 

[Blues_MTA]

So, du = Sec(x)^2+tan(x)Sec(x)/Sec(x)+Tan(x) dx, reduces to sec(x)dx?

 

[micromass]

So, du = Sec(x)^2+tan(x)Sec(x)/Sec(x)+Tan(x) dx, reduces to sec(x)dx?

Yes, just factor sec(x) in the numerator.

 

[Blues_MTA]

Ah! Thank you, I've been looking at it for so long i didnt even see that! the only thing left is that i don't understand how the integral of 1/(u^1/2) is 2(u^1/2)

 

[micromass]

Ah! Thank you, I've been looking at it for so long i didnt even see that! the only thing left is that i don't understand how the integral of 1/(u^1/2) is 2(u^1/2)

In, general, what is the integral of uⁿ?? You must have seen a special formula for that...

 

[Blues_MTA]

HaHa...wow I see it all now, I need to sleep, Thank you so much!