# Ali's question at Yahoo! Answers regarding an indefinite integral

#### MarkFL

Staff member
Here is the question:

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

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

#### MarkFL

Staff member
Hello Ali,

We are given to evaluate:

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

I would first use the substitution:

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

and we now have:

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

Now, using integration by parts, I would let:

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

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

and now we have:

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

Next, using the substitution:

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

we may write:

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

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

Back-substitute for $u$:

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

Back-substitute for $w$:

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