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I am having trouble figuring this out.

$$\int_0^{1/3} \sec^3(\pi x) \, dx$$

We are currently doing integration by parts,, so I set $g(x)=\sec^3(\pi x)$ and $f'(x)=1$. I arrived at: $$x\sec^3(\pi x) - 3\pi \int x\tan(\pi x)\sec^3(\pi x) dx$$ I am lost here, I tried u substitution for $\tan(\pi x)$ so that I can get rid of $\sec^2(\pi x)$ but that doesnt help.

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marked as duplicate by Gerry Myerson, Ross Millikan, Henry T. Horton, Asaf Karagila, 5PM Feb 11 '13 at 2:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

$\int\sec^3x\,dx$ is a well-known hard example, but I'm sure it has been done on this site before. Found it ---… – Gerry Myerson Feb 11 '13 at 1:55
up vote 3 down vote accepted

Part of the problem is that your first integration by parts is a poor choice. Let $u=\sec\pi x$ and $dv=\sec^2\pi x \, dx$, so that $du=\pi\sec\pi x\tan\pi x \, dx$ and $v=\frac1{\pi}\tan\pi x$. Now your (indefinite) integral $-$ call it $I$ $-$ is

$$I=\frac1{\pi}\sec\pi x\tan\pi x-\int\sec\pi x\tan^2\pi x \, dx\;.\tag{1}$$

Apply a trig identity to write that last integral as

$$\int\sec\pi x\left(\sec^2\pi x-1\right)dx=I-\int\sec\pi x \, dx$$

and substitute into $(1)$ to get

$$I=\frac1{\pi}\sec\pi x\tan\pi x-I+\int\sec\pi x \, dx\;.\tag{2}$$

Now solve $(2)$ for $I$; the remaining integral is one that you should know.

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This is a recurring question. See this article:

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It can be shown that (as pointed out in the comment of Michael Hardy) $$\int \sec^3 \theta d\theta=\frac{\sec\theta \tan\theta}{2}+\frac{\ln|\sec\theta+\tan\theta|}{2} +c.$$

Going back to your integral, we let $\theta=\pi x$ and we get $$F(x)=\int \sec^3(\pi x) \, dx =\frac{\sec(\pi x) \tan(\pi x)} {2\pi}+\frac{\ln|\sec(\pi x) +\tan(\pi x)|}{2\pi}+c.$$Thus, $$\int_0^{1/3} \sec^3(\pi x) \, dx=F(1/3)-F(0)$$ and we are done.

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