MATH105 April 2011
• Q1 (a) • Q1 (b) • Q2 • Q3 • Q4 • Q5 (a) • Q5 (b) • Q5 (c) • Q6 • Q7 • Q8 (a) • Q8 (b) • Q8 (c) • Q9 (a) • Q9 (b) • Q9 (c) • Q9 (d) • Q9 (e) • Q9 (f) • Q9 (g) •
Question 01 (b)

Find the value of the following definite integral:
 $\int _{0}^{\pi /2}\cos ^{3}(x)\sin ^{2}(x)\,dx$

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Hint

The power of $\cos \,x$ is odd. Is there a strategy that would be useful?

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Solution

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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
 ${\begin{aligned}\int \cos(x)(1\sin ^{2}(x))\sin ^{2}(x)\,dx&=\int \cos(x)(1u^{2})u^{2}\,{\frac {du}{\cos(x)}}\\&=\int (1u^{2})u^{2}\,du\\&=\int u^{2}u^{4}\,du\\&={\frac {u^{3}}{3}}{\frac {u^{5}}{5}}+C.\end{aligned}}$
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:
 ${\begin{aligned}\int _{0}^{\pi /2}\cos ^{3}(x)\sin ^{2}(x)\,dx&=\left({\frac {\sin ^{3}(x)}{3}}{\frac {\sin ^{5}(x)}{5}}\right.{\Bigr }_{0}^{\pi /2}\\&=\left({\frac {\sin ^{3}(\pi /2)}{3}}{\frac {\sin ^{5}(\pi /2)}{5}}\right)\left({\frac {\sin ^{3}(0)}{3}}{\frac {\sin ^{5}(0)}{5}}\right)\\&=\left({\frac {1}{3}}{\frac {1}{5}}\right)\left({\frac {0}{3}}{\frac {0}{5}}\right)\\&={\frac {2}{15}}.\end{aligned}}$

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