Integrals Involving Trig Functions — Question 5

Question 5 – Solution

Separate one factor of $\sec(x)\tan(x)$: $$\sec^5(x)\tan^3(x) =\sec^4(x)\tan^2(x)\bigl(\sec(x)\tan(x)\bigr).$$ Use $\tan^2(x)=\sec^2(x)-1$ to rewrite the remaining tangent power: $$\sec^4(x)\tan^2(x) =\sec^4(x)\bigl(\sec^2(x)-1\bigr).$$ Now let $$u=\sec(x), \qquad du=\sec(x)\tan(x)\,dx.$$ The integral becomes $$\begin{align*} \int u^4(u^2-1)\,du &=\int(u^6-u^4)\,du\\ &=\frac{u^7}{7}-\frac{u^5}{5}+C. \end{align*}$$ Substituting $u=\sec(x)$ gives $$\boxed{\displaystyle \int\sec^5(x)\tan^3(x)\,dx =\frac17\sec^7(x)-\frac15\sec^5(x)+C}.$$ Saving $\sec(x)\tan(x)$ is useful because it supplies $du$, while the identity converts all remaining factors into powers of $u$.