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I need help on these three integrals. Any hints on which method to use are greatly appreciated.

$$1)\ \int \frac{1}{\cos^4 x}\tan^3 x\mathrm{d}x$$ $$2)\ \int \frac{1}{\sin 2x}(3\cos x + 7\sin x)\mathrm{d}x$$ $$3)\ \int \frac{1}{\sqrt{x}}\sin^3 (3\sqrt{x})\mathrm{d}x$$

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Just out of curiosity, where do the list come from? (if it's from a book, which one?) – Jean-Claude Arbaut Aug 22 '14 at 19:33
In #1, you can write the integrand as $\tan^{3}x\sec^{4}x=\tan^{3}x(1+\tan^{2}x)\sec^{2}x$ and then let $u=\tan x$. – user84413 Aug 22 '14 at 19:39
The list is a review a professor gave me for Calc 3. Just reviewing some Calc 2 topics. – Jackson Aug 22 '14 at 20:07
1 question per question, please. :) – apnorton Aug 28 '14 at 3:21
up vote 8 down vote accepted

1) The integrand is the same as $\large\frac{\sin^3(x)}{\cos^7(x)} = \frac{(1-\cos^2(x)) \sin(x)}{\cos^7(x)}$. That reduces to two terms which can both be turned into $\large\frac{dz}{z^n}$ by a simple substitution.

2) Note $\sin(2 x) = 2 \sin(x) \cos(x)$. With that you simply have an integral of $\sec(x)$ plus an integral of $\csc(x)$.

3) Try the substitution $z = \sqrt{x}$. That should turn the integral into the form $\sin^3(3 z) = \sin(3 z) (1 - \cos^2(3 z))$. That should be doable by a simple substitution.

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Yes that's right, thanks. I edited it. – Bridgeburners Aug 22 '14 at 19:35
Thanks a bunch, i don't know why these looked so hard to me. I guess you have to know the tricks. I haven't done Calc 2 stuff in a while so I'm reviewing. Thanks again! – Jackson Aug 22 '14 at 19:58

Just because it wasn't mentioned explicitly I think it is worth mentioning on number 1. The integrand is $\tan^3x\sec^4x$ which is equal to

$$ \tan^3x(1+\tan^2x)\sec^2x$$

This is easily integrated as $\sec^2x$ is the derivative of $\tan x$.

$$ \tan^3x\sec^2x+\tan^5x\sec^2x $$

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