Integrals Involving Trig Functions — Question 1

Question 1 – Solution

The power of sine is odd, so separate one sine factor: $$\sin^5(x)=\sin^4(x)\sin(x).$$ The remaining even power can be converted using $\sin^2(x)=1-\cos^2(x)$: $$\sin^4(x)=\bigl(1-\cos^2(x)\bigr)^2.$$ Thus, $$\int\sin^5(x)\cos^4(x)\,dx =\int\bigl(1-\cos^2(x)\bigr)^2\cos^4(x)\sin(x)\,dx.$$ Let $$u=\cos(x),\qquad du=-\sin(x)\,dx.$$ Then the integral becomes $$\begin{align*} -\int(1-u^2)^2u^4\,du &=-\int(1-2u^2+u^4)u^4\,du\\ &=-\int(u^4-2u^6+u^8)\,du\\ &=-\frac{u^5}{5}+\frac{2u^7}{7}-\frac{u^9}{9}+C. \end{align*}$$ Substituting $u=\cos(x)$ gives $$\boxed{\displaystyle \int\sin^5(x)\cos^4(x)\,dx =-\frac{\cos^5(x)}5+\frac{2\cos^7(x)}7-\frac{\cos^9(x)}9+C}.$$ Separating one sine factor is efficient because it supplies the differential $-\sin(x)\,dx$ while every remaining factor becomes a polynomial in $u$.