# Math Help - Integral Help

1. ## Integral Help

Thanks!!

2. This is really easy.

Have you tried $u=\tan x$?

3. ## Re:

Yea I agree. I just worked some really hard ones, then they stuck in this easy one. Kind of like playing with indeterminate limits then all of a sudden coming across one that converges by just plugging in the numbers.

Thanks,
-qbkr21

4. Originally Posted by qbkr21
Yea I agree. I just worked some really hard ones, then they stuck in this easy one. Kind of like playing with indeterminate limits then all of a sudden coming across one that converges by just plugging in the numbers.

Thanks,
-qbkr21
Use $u=tanx$ $\rightarrow$ $du=sec^2x\,dx$

$
\int tan^3x\,sec^2x\,dx = \int u^3\,du = \dfrac{1}{4}\,u^4\,+C = \dfrac{1}{4}\,tan^4x\,+C
$