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

Thanks so much in advance!

Question

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

Thanks so much in advance!

anonymous · Answer

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

gecko · Answer

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 (:

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

gecko · Answer

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

anonymous · Answer

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

anonymous · Answer

upper minus lower yes?

gecko · Answer

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

anonymous · Answer

yw