Question

Integrate from 0 to pi/8: (sec^2(2x))/(tan(2x)+1)

Thanks so much in advance!

Answer 1

put
\[u=\tan(2x)+1\]
\[du=2\sec^2(2x)dx\]

Answer 2

O: Ahhh I think I've got it lol. Wasn't sure if it was u-sub since we just started with it in class.. Thanks satellite! I believe this isn't the first time you've saved me (:

Answer 3

\[u(0)=\tan(0)+1=1\]
\[u(\frac{\pi}{8})=\tan(\frac{\pi}{4})+1=2\]

Answer 4

yw. i think the result will be
\[\int_1^2\frac{1}{2}\frac{1}{u}du=\frac{1}{2}\ln(2)\]

Answer 5

or even
\[\ln(\sqrt{2})\] if you prefer

Answer 6

Thank you much (: Would it be -(1/2)ln(2) since you subtract with the upper bound?

Answer 7

oh no. it is F(b)-F(a)

Answer 8

upper minus lower yes?

Answer 9

Ohh derp :P You're right. I wrote it down wrong lol. Thanks!

Answer 10

yw