Integrals Involving Trig Functions — Question 4

Question 4 – Solution

The factor $\sec^2(x)\,dx$ is exactly the differential of $\tan(x)$, so let $$u=\tan(x), \qquad du=\sec^2(x)\,dx.$$ Then $$\begin{align*} \int\tan^7(x)\sec^2(x)\,dx &=\int u^7\,du\\ &=\frac{u^8}{8}+C. \end{align*}$$ Substitute $u=\tan(x)$: $$\boxed{\displaystyle \int\tan^7(x)\sec^2(x)\,dx=\frac18\tan^8(x)+C}.$$ As a check, $$\begin{align*} \frac{d}{dx}\left[\frac18\tan^8(x)\right] &=\frac18\cdot8\tan^7(x)\sec^2(x)\\ &=\tan^7(x)\sec^2(x), \end{align*}$$ which is the original integrand.