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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

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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 do not recognize any of those subsitutions, where are they from? Is that a tan half or double angle?2012-06-03
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    No trigonometric identity except for $\sec^2 x = 1 + \tan^2 x$ and $(\tan x)' = \sec^2 x$ is used here.2012-06-03
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    I do not understand the $tan^2 x / 2sec^4 x$2012-06-03
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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. Rather, this shows that there are many ways 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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    I do not understand how I was suppose to see that this was the right way to do the problem.2012-06-03
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    @Jordan: *A lot* (if not most) of these things come with practice. *A LOT* of practice. I am talking about dozens of exercises, literally dozens. I can assure you that after solving 50-60 of these questions it will be a lot easier.2012-06-03
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    @AsafKaragila I put in 30 hours already.2012-06-03
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    @Jordan: I had to put several months into studying some things. 30 hours is not *that* much.2012-06-03
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    @AsafKaragila I do not understand how you had time to put months in, I have 6 more sections to go and I spent 30 hours on one, and I only have 24 hours if I do not sleep to get this done.2012-06-03
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    @Jordan: We study in different ways, apparently. If you have that little time my only suggestion is that you do not stress yourself (mentally) too much. Stress reduces your learning abilities by half and makes you ineffective (not to mention wasting time feeling despair instead of studying). Good luck.2012-06-03
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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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    I did do that but it just seems to make the problem more complicated and then there is no u substitution that will reduce the problem.2012-06-03
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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, $$I = \int \left( \dfrac{t^2 - 1}{t} \right) dt = \dfrac{t^2}{2} - \log(t) + C = \dfrac{\cos^2(x)}{2} - \ln(\lvert\cos(x) \rvert) + C$$

......