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

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MarkFL

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Feb 24, 2012
13,775
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.
 
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MarkFL

Administrator
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Feb 24, 2012
13,775
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 $w$:

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