Question:
Find the value of the expression $$\int \tan^2\theta \sec^4\theta\ d\theta$$ by using the substitution $u=\tan\theta$.
My Working:
If $u=\tan\theta$, then
\begin{align} \frac{du}{d\theta}&=\sec^2\theta\\ du&=\sec^2\theta\cdot d\theta\\ \int\tan^2\theta \sec^4\theta\ d\theta&=\int \tan^2\theta\cdot \sec^2\theta\cdot(\sec^2\theta\cdot d\theta)\\ &=^?\int u^2\cdot u^\prime\cdot du \end{align}
After this, I do not know how to proceed as there is an extra remaining factor of $\sec^2\theta$, and I cannot use the product to sum formula as there is no $\sin x$ or $\cos x$. Could anyone please help? Thanks!
\begin{align} \int\tan^2\theta \sec^4\theta\ d\theta&=\int \tan^2\theta\cdot \sec^2\theta\cdot(\sec^2\theta\cdot d\theta)\qquad(\text{say, }\tan\theta=u)\\ &=\int u^2\cdot (1+u^2)\cdot du\\ &=\int u^2 du +\int u^4 du\\ &=\frac{u^3}{3}+\frac{u^5}{5}+c\qquad(\text{some constant, }c)\\ &=\frac{\tan^3\theta}{3}+\frac{\tan^5\theta}{5}+c \end{align}