Evaluate: $\int \tan^5 (x). \sec (x) dx$

Question

Evaluate: $\int \tan^5 (x). \sec (x) dx$

My Attempt: $$=\int \tan^5 (x).\sec (x).dx$$ $$=\int \tan^4 (x).\sec (x).\tan (x) dx$$ $$=\int (\sec^2 (x)-1)^2 .\sec (x).\tan (x) dx$$ $$=\int (\sec^4 (x)-2\sec^2 (x)+1).\sec (x).\tan (x) dx$$

Answer 1

It usually helps to convert to $sin(x)$ and $cos(x)$.

$$\begin{aligned}\int \tan^5(x)\sec(x)dx &= \int \frac{\sin^5(x)}{\cos^6(x)}dx \\ &=-\int \frac{(1-\cos^2(x))^2}{\cos^6(x)}(-\sin(x))dx \\ &=-\int \frac{(1-u^2)^2}{u^6}du \end{aligned}$$ where $u=\cos (x)$. I'll leave the remaining steps to you since they're fairly easy.