Science:Math Exam Resources/Courses/MATH105/April 2011/Question 01 (b)/Solution 1

We will first compute the indefinite integral and evaluate our results at the boundaries of the integral to get the final result.

The first step is to split a $\cos(x)$-factor from $\cos ^{3}(x)$ in the integrand:

$\int \cos ^{3}(x)\sin ^{2}(x)\,dx=\int \cos(x)\cos ^{2}(x)\sin ^{2}(x)\,dx.$

Now we use the trigonometric Pythagoras to write $\cos ^{2}(x)=1-\sin ^{2}(x)$ and arrive at

$\int \cos(x)\cos ^{2}(x)\sin ^{2}(x)\,dx=\int \cos(x)(1-\sin ^{2}(x))\sin ^{2}(x)\,dx.$

The next step is to use the substitution rule with $u=\sin(x)$. This implies $du/dx=\cos(x)$ or equivalently $dx=du/\cos(x)$. Substituting this into the integral yields

${\begin{aligned}\int \cos(x)(1-\sin ^{2}(x))\sin ^{2}(x)\,dx&=\int \cos(x)(1-u^{2})u^{2}\,{\frac {du}{\cos(x)}}\\&=\int (1-u^{2})u^{2}\,du\\&=\int u^{2}-u^{4}\,du\\&={\frac {u^{3}}{3}}-{\frac {u^{5}}{5}}+C.\end{aligned}}$

Rewriting this integral in terms of $x$ by using the substitution equation $u=\sin(x)$ results in

$\int \cos ^{3}(x)\sin ^{2}(x)\,dx={\frac {\sin ^{3}(x)}{3}}-{\frac {\sin ^{5}(x)}{5}}+C.$

Therefore, we can compute our original definite integral as follows:

${\begin{aligned}\int _{0}^{\pi /2}\cos ^{3}(x)\sin ^{2}(x)\,dx&=\left({\frac {\sin ^{3}(x)}{3}}-{\frac {\sin ^{5}(x)}{5}}\right.{\Bigr |}_{0}^{\pi /2}\\&=\left({\frac {\sin ^{3}(\pi /2)}{3}}-{\frac {\sin ^{5}(\pi /2)}{5}}\right)-\left({\frac {\sin ^{3}(0)}{3}}-{\frac {\sin ^{5}(0)}{5}}\right)\\&=\left({\frac {1}{3}}-{\frac {1}{5}}\right)-\left({\frac {0}{3}}-{\frac {0}{5}}\right)\\&={\frac {2}{15}}.\end{aligned}}$