Answer
$\frac{1}{3}\tan^3\theta+\frac{1}{5}\tan^5\theta+C$
Work Step by Step
$\int\tan^2\theta\sec^4\theta\ d\theta$
Since the power of secant is even, we save a factor of $\sec^2\theta$, and express the rest in terms of $\tan\theta$:
$=\int\tan^2\theta\sec^2\theta\sec^2\theta\ d\theta$
$=\int\tan^2\theta(1+\tan^2\theta)\sec^2\theta\ d\theta$
Let $u=\tan\theta$. Then $du=\sec^2\theta\ d\theta$.
$=\int u^2(1+u^2)du$
$=\int(u^2+u^4)du$
$=\frac{1}{3}u^3+\frac{1}{5}u^5+C$
$=\boxed{\frac{1}{3}\tan^3\theta+\frac{1}{5}\tan^5\theta+C}$
