Question

IBP:

Answer 1

\[\LARGE \int t~sec^2(2t)~dt\]

Answer 2

\[\Large\rm \int\limits \sec^2(2t)dt\]

Answer 3

Like do you know how to deal with the inner 2, without making an extra substitution?

Answer 4

blah my message didn't post,
it said:
"quick question: can you do this integral in your head?"

Answer 5

Would it be \(\Large \frac{1}{2}tan(2t)+C\)?

Answer 6

Yes, good.
We end up with a 1/2 because of the 2.

Answer 7

So that takes care of our dv and v.\[\Large\rm dv=\sec^2(2t), \qquad\qquad v=\frac{1}{2}\tan(2t)\]

Answer 8

Err I should be a little more accurate and say:\[\Large\rm dv=\sec^2(2t)dt\]

Answer 9

\[\Large\rm u=t, \qquad\qquad \quad dv=\sec^2(2t)dt\]\[\Large\rm du=dt,\qquad\qquad v=\frac{1}{2}\tan(2t)\]

Answer 10

Understand why I chose to for the u?

Answer 11

Yup, yup~

Answer 12

So where ya stuck? >:U
Just do it!!

Answer 13

Just to clarify, after all that it would end up as
\[\LARGE \frac{t}{2}tan(2t) -\frac{1}{2} \int tan(2t)dt\]?

Answer 14

Mmmmmm ya looks good.

Answer 15

Remember how to integrate tangent?
That one's not too bad.

Answer 16

Uhm, \(\LARGE \int~tanx~dx=ln|sec~x|+C\)?

Answer 17

Mmm ya that sounds right.

Answer 18

Okay, thank you zepdrix :D