Question

How do you evaluate the integral of tan^3x sec^2x dx?

Answer 1

try
\[u=\tan(x)\]

Answer 2

Correction. Its the integral of (tan^3)x(sec^2)xdx

Answer 3

then
\[du=\sec^2(x)dx\] giving
\[\int u^3du\]

Answer 4

I know you break it into this: (tan^2)x(sec^2)xtanxdx

Answer 5

1/4 tan^4x +C

Answer 6

no don't break it. it is a straight forward u - sub

Answer 7

That's what my teacher wants.