MHB Ali's question at Yahoo Answers regarding an indefinite integral

[MarkFL]

Here is the question:

What is the integral of asec(x^(1/2))?

Please show your work.

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Ali,

We are given to evaluate:

$$I=\int\sec^{-1}\left(\sqrt{x} \right)\,dx$$ where $$0\le x$$

I would first use the substitution:

$$w=\sqrt{x}\,\therefore\,dx=2w\,dw$$

and we now have:

$$\int 2w\sec^{-1}(w)\,dw$$

Now, using integration by parts, I would let:

$$u=\sec^{-1}(w)\,\therefore\,du=\frac{1}{w\sqrt{w^2-1}}\,dw$$

$$dv=2w\,dw\,\therefore\,v=w^2$$

and now we have:

$$I=w^2\sec^{-1}(w)-\int\frac{w}{\sqrt{w^2-1}}\,dw$$

Next, using the substitution:

$$u=w^2-1\,\therefore\,du=2w\,dw$$

we may write:

$$I=w^2\sec^{-1}(w)-\frac{1}{2}\int u^{-\frac{1}{2}}\,du$$

$$I=w^2\sec^{-1}(w)-u^{\frac{1}{2}}+C$$

Back-substitute for $u$:

$$I=w^2\sec^{-1}(w)-\sqrt{w^2-1}+C$$

Back-substitute for $w$:

$$I=x\sec^{-1}\left(\sqrt{x} \right))-\sqrt{x-1}+C$$