Integrals Involving Trig Functions — Question 8

Question 8 – Solution

Separate one sine factor and use $\sin^2(x)=1-\cos^2(x)$: $$\sin^3(x)=\sin(x)\bigl(1-\cos^2(x)\bigr).$$ Then $$\int\frac{\sin^3(x)}{\cos^4(x)}\,dx =\int\frac{1-\cos^2(x)}{\cos^4(x)}\sin(x)\,dx.$$ Let $$u=\cos(x), \qquad du=-\sin(x)\,dx.$$ The integral becomes $$\begin{align*} -\int\frac{1-u^2}{u^4}\,du &=-\int(u^{-4}-u^{-2})\,du\\ &=-\left(-\frac13u^{-3}+u^{-1}\right)+C\\ &=\frac13u^{-3}-u^{-1}+C. \end{align*}$$ Substituting $u=\cos(x)$ gives $$\boxed{\displaystyle \int\frac{\sin^3(x)}{\cos^4(x)}\,dx =\frac{1}{3\cos^3(x)}-\frac{1}{\cos(x)}+C}.$$ Equivalently, the answer is $\frac13\sec^3(x)-\sec(x)+C$ on intervals where the original integrand is defined.