Question

Meh another trig integration problem. Thought I'd get these by now :/

Answer 1

\[\int\limits_{}^{}8(\tan^2x+\tan^4x)dx\]

Answer 2

My first instinct is to replace tan^2x with 1-sec^2x lol

Answer 3

Try factoring first!

Answer 4

ok\[8\int\limits_{}^{}\tan^2x(1+\tan^2x)\]

Answer 5

then?

Answer 6

You can now use your reflex! lol

Answer 7

tan^2x(1+(1-sec^2x)) ....

Answer 8

hmmm that's not good

Answer 9

I thought 1+tan^2=sec^2 ???

Answer 10

ok so now I have \[8\int\limits_{}^{}\tan^2xsec^2x dx\]

Answer 11

u= tan

du= sec^2x

Answer 12

Exactly, you're on your way to success! Go! Go! Go!

Answer 13

hehe \[8\int\limits_{}^{}u^2du\]

Answer 14

Now you can finish it off with your eyes closed! lol

Answer 15

\[\frac{ 8\tan^3x }{ 3 }+ C\]

Answer 16

yass

Answer 17

There you go, man!

Answer 18

Thanks! Such a relief. I was starting to panic because I'm taking calc 2 right now summer term and I've fallen behind. :S but I feel like I'm catching up and actually starting to understand this.

Answer 19

I should actually be studying sequences and series right now. My class actually already went over techniques of integration :s

Answer 20

Nice work!

Answer 21

:)