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

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. – AD. Jun 3 '12 at 17:17

$$I = \int \cos^2(x) \tan^3(x) dx = \int \cos^2(x) \dfrac{\sin^3(x)}{\cos^3(x)} dx = \int \dfrac{\sin^3(x)}{\cos(x)} dx$$ $$I = \int \dfrac{\sin^3(x)}{\cos(x)} dx = \int \dfrac{\sin(x)}{\cos(x)} \sin^2(x) dx = \int \dfrac{\sin(x)}{\cos(x)} \left( 1 - \cos^2(x)\right) dx$$ $$I = \int \dfrac{\left(\cos^2(x)-1\right)}{\cos(x)} (-\sin(x)) dx$$

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

......

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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. – user138246 Jun 3 '12 at 16:48
@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. – Asaf Karagila Jun 3 '12 at 16:51
@AsafKaragila I put in 30 hours already. – user138246 Jun 3 '12 at 16:52
@Jordan: I had to put several months into studying some things. 30 hours is not that much. – Asaf Karagila Jun 3 '12 at 16:54
@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. – user138246 Jun 3 '12 at 16:55

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. – user138246 Jun 3 '12 at 16:38
@Jordan Please publish details on what you have tried, perhaps someone can help you to put the finger on the real issue. – AD. Jun 3 '12 at 16:41

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? – user138246 Jun 3 '12 at 16:47
No trigonometric identity except for $\sec^2 x = 1 + \tan^2 x$ and $(\tan x)' = \sec^2 x$ is used here. – Sangchul Lee Jun 3 '12 at 16:49
I do not understand the $tan^2 x / 2sec^4 x$ – user138246 Jun 3 '12 at 16:52
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. – Sangchul Lee Jun 3 '12 at 16:58