Int [ sec^2(sqrt(x)) / sqrt(x) ] dx (which method to use?)

[savemordecai]

Here's the problem :

Integral [ (sec^2(sqrt(x))) / (sqrt(x)) dx]

I'm just not sure where to start out on this problem.
Any tips or hints would be greatly appreciated, thanks.

 

[soroban]

Re: Integration (not sure which method to use).

Hello, savemordecai!

\(\displaystyle \L\int \frac{\sec^2(\sqrt{x})}{\sqrt{x}}\,dx\)

You have: \(\displaystyle \L\:\int\)\(\displaystyle x^{-\frac{1}{2}}\sec^2\left(x^{\frac{1}{2}}\right)\,dx\)

\(\displaystyle \text{Let }u\,=\,x^{\frac{1}{2}}\)

 

[savemordecai]

Re: Integration (not sure which method to use).

Hello, savemordecai!

\(\displaystyle \L\int \frac{\sec^2(\sqrt{x})}{\sqrt{x}}\,dx\)

You have: \(\displaystyle \L\:\int\)\(\displaystyle x^{-\frac{1}{2}}\sec^2\left(x^{\frac{1}{2}}\right)\,dx\)

\(\displaystyle \text{Let }u\,=\,x^{\frac{1}{2}}\)

So du=1/2x^-1/2 dx
And that becomes : 2*Integral[Sec^2(u)du]
Which would give 2tanu
And substituting back in gives an answer of 2tan(sqrt[x]) + c

Thanks for the help.
I have a nasty problem of making problems way more complicated than they are.