Question

Calculus II question

Answer 1

\[\int\limits_{}^{}\sec^2xtanxdx \]

Answer 2

I know I'll use a u substiturion, I just dont know how to do it. a walk through would be much appreciated.

Answer 3

ok, so
let tanx=t, t is some variable varying along with tanx.
take differentials on both the sides. what do you get?

Answer 4

awwww! I got it t =tanx dt = secxtanx so the new integral is \[\int\limits_{}^{}u du \]

Answer 5

err t

Answer 6

final answer is \[\frac{ 1 }{ 2 } \tan^2x +C\]

Answer 7

\[dt=\sec ^{2}xdx.\]
yes, your integration is correct.

Answer 8

Awesome, thanks man. @Abhishek619

Answer 9

sure! anytime! :)