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I have no idea how to do a problem like this. I know I can't do $u$ substitution because $\tan$ or $\sec$ doesn't cancel out both the $\tan$ and the $\cos$.

$\int \cos^2 (x) \tan^3 (x) dx$

$\int \cos^2 (x) (\sec^2 (x) - 1) \tan (x) dx$

From here I can't really do anything because no u will cancel out everything.

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    Please note that this is homework. Jordan will not gain anything from a complete solution.2012-06-03

4 Answers 4

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Just observe that

$ \begin{align*} \cos^2 x \, \tan^3 x & = \frac{\tan^3 x}{\sec^2 x} \\ & = \frac{\tan^2 x \cdot \tan x}{\sec^2 x} \cdot \frac{2 \sec^2 x}{2 \sec^2 x} \\ & = \frac{\tan^2 x}{2 \sec^4 x} (2 \tan x \, \sec^2 x)\\ & = \frac{\tan^2 x}{(1+\tan^2 x)^2} (\tan^2 x)', \end{align*}$

thus the substitution $t = \tan^2 x$ gives

$\int \cos^2 x \, \tan^3 x \, dx = \int \frac{t}{2(1+t)^2} \; dt.$

Now the rest is clear.

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    I added some intermediate steps. Hope now the underlying idea became clearer... I do not think this is just the fastest and easiest way. $R$ather, this shows that there are man$y$ wa$y$s to calculate an integral.2012-06-03
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$\int \cos^2 x \tan^3 x\ dx = \int \frac{\sin^3 x}{\cos x}\ dx = \int \frac{1-\cos^2 x}{\cos x}\sin x\ dx$

$u=\cos x$ substitution

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    @AsafKaragila I just do not understand how I am not suppose to be panicking and worrying about this. I just can not see how I have the time to catch up when I have 6 more sections to do that get even harder when I have spent 5 days trying to learn 2 sections and failing and I have 1 day to learn the other 6.2012-06-03
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Hint:

  1. $\tan x=...$

  2. $\sin^2x=1-...$

  3. Make a substitution!

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    @Jordan Please publish details on what you have tried, perhaps someone can help you to put the finger on the real issue.2012-06-03
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$I = \int \cos^2(x) \tan^3(x) dx = \int \cos^2(x) \dfrac{\sin^3(x)}{\cos^3(x)} dx = \int \dfrac{\sin^3(x)}{\cos(x)} dx$ $I = \int \dfrac{\sin^3(x)}{\cos(x)} dx = \int \dfrac{\sin(x)}{\cos(x)} \sin^2(x) dx = \int \dfrac{\sin(x)}{\cos(x)} \left( 1 - \cos^2(x)\right) dx$ $I = \int \dfrac{\left(\cos^2(x)-1\right)}{\cos(x)} (-\sin(x)) dx$

Now let $\cos(x) = t$. This gives us $-\sin(x) dx = dt$ Hence, $I = \int \left( \dfrac{t^2 - 1}{t} \right) dt = \dfrac{t^2}{2} - \log(t) + C = \dfrac{\cos^2(x)}{2} - \ln(\lvert\cos(x) \rvert) + C$

......