Question

I need help with this problem: integrate sec^2(pix)tan^5(pix) dx

Answer 1

\[\int\sec^2\pi x\tan^5\pi x~dx\]
Let \(u=\tan\pi x\), so that \(du=\pi\sec^2\pi x~dx\), or \(\dfrac{1}{\pi}du=\sec^2\pi x~dx\):
\[\frac{1}{\pi}\int u^5~du\]

Answer 2

I first tried making u=tan(pix)

Answer 3

ohhh, I messed up pretty early on. I didn't take out my constant

Answer 4

Yep, gotta watch out for that chain rule.

Answer 5

yeah, I tend to miss that :/

Answer 6

then you just integrate u and sub tan(pix) back in

Answer 7

Right

Answer 8

\[\frac{ 1 }{6 \pi }\tan ^{6}(pix)\]

Answer 9

thank you!

Answer 10

You're welcome!