Integral of $\sec^2{x} \tan^2{x}$ I'm once again stuck; I'm trying to find $$\int{(\sec^2{x} \tan^2{x})dx}$$ but end up with things like: $\tan^2x-\int{2\tan^2x \sec^2x}$, which doesn't help.
Would it be best to approach using integration by parts or substitution?
 A: Let $u=\tan x$, $du=\sec^2 x~dx$. The integral becomes:
$$\int u^2\ du$$
Can you continue from here?
A: I'm assuming you got your answer using 'by parts'. Below is how you could finish it using 'by parts'. Note that other solutions such as by @ZacharySelk are simpler.
Using your line of working:
$$\int sec^2x \tan^2x dx = tan^2x - 2\int \sec^2x \tan^2x dx$$
You can move the $- 2\int \sec^2x \tan^2x dx$ to the left hand side of the equation by addition.
$$\int \sec^2x \tan^2x dx+ 2\int \sec^2x \tan^2x dx= tan^2x +c, c\in\mathbb{R}$$
Note that once we have a side without an integral on it you need to include a constant of integration. I have used $c$.
The two expressions on the left hand side are the same so you can add them giving:
$$3\int \sec^2x \tan^2x dx= tan^2x +c$$
So simply divide by 3 to get your answer:
$$\int \sec^2x \tan^2x dx= \frac{tan^2x}{3} +\frac{c}{3}$$
Note that as $c$ is an real number we could replace $\frac{c}{3}$ with $c_2$ to write the answer more neatly as:
$$\int \sec^2x \tan^2x dx= \frac{tan^2x}{3} +c_2$$
A: Putting $u = \tan x$, $du=\sec^2x\ dx$, we get:
$\ \ \ \ \int u^2du$
$=\frac{u^3}{3}+c$
$=\frac{\tan^3x}{3}+c$
