Science:Math Exam Resources/Courses/MATH101/April 2017/Question 05 (a)/Solution 1

Recall $(\cos(x))'=-\sin(x)$ and the Pythagorean Identity $\cos ^{2}(x)+\sin ^{2}(x)=1$.

Consider the substitution $u=\cos(x)$. We observe that as $x$ ranges from $0$ to $\pi$, the function $u$ ranges from $1$ to $-1$.

Moreover, $du=-\sin(x)\,dx$ and by the Pythagorean Identity,

$\sin ^{5}(x)=\sin ^{4}(x)\sin(x)=(1-\cos ^{2}(x))^{2}\sin(x).$

Plugging this into the integral and applying the substitution, we get

$\int _{0}^{\pi }\sin ^{5}(x)dx=\int _{0}^{\pi }(1-\cos ^{2}(x))^{2}\sin(x)\,dx=-\int _{1}^{-1}(1-u^{2})^{2}\,du=\int _{-1}^{1}(1-u^{2})^{2}\,du=2\int _{0}^{1}(1-u^{2})^{2}\,du.$

The last equality follows from that $(1-u^{2})^{2}$ is an even function.

Finally, we evaluate the integral as follows. $\int _{0}^{\pi }\sin ^{5}(x)dx=2\int _{0}^{1}1-2u^{2}+u^{4}\,du=2\left[u-{\frac {2}{3}}u^{3}+{\frac {1}{5}}u^{5}\right]_{u=0}^{u=1}=\color {blue}2\left(1-{\frac {2}{3}}+{\frac {1}{5}}\right)$