Question

How would I start off this integral?

Answer 1

\[\LARGE \int cos^2x~tan^3x~dx\]

Answer 2

He's the odd man out

Answer 3

tanx=sinx/cosx

Answer 4

hint: tan^2x = sin^2x / cos^2x

Answer 5

Oh I think I see it finally.\[\Large\rm \int\limits \cos^2x \frac{\sin^3x}{\cos^3x}~dx\]\[\Large\rm \int\limits \frac{\sin^3x}{\cos x}~dx\]\[\Large\rm \int\limits \frac{1-\cos^2x}{\cos x}~(\sin x ~dx)\]Then let u=cos x

Answer 6

Kind of a tricky one! :O

Answer 7

\[\int\limits \frac{\sin^3 x}{\cos x}dx\] \[\cos x = u\]\[\sin x \ \ dx = -du\]\[u^2=\cos^2x=1-\sin^2x\]\[\sin^2x=1-u^2\]
\[\int\limits \frac{(1-u^2)(-du)}{u}\]

Answer 8

Thank you guys :)