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

MATH105 April 2011

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Question 01 (b)
Find the value of the following definite integral: $\int _{0}^{\pi /2}\cos ^{3}(x)\sin ^{2}(x)\,dx$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: ${\begin{aligned}\int _{0}^{\pi /2}\cos ^{3}(x)\sin ^{2}(x)\,dx&=\left({\frac {\sin ^{3}(x)}{3}}-{\frac {\sin ^{5}(x)}{5}}\right.{\Bigr |}_{0}^{\pi /2}\\&=\left({\frac {\sin ^{3}(\pi /2)}{3}}-{\frac {\sin ^{5}(\pi /2)}{5}}\right)-\left({\frac {\sin ^{3}(0)}{3}}-{\frac {\sin ^{5}(0)}{5}}\right)\\&=\left({\frac {1}{3}}-{\frac {1}{5}}\right)-\left({\frac {0}{3}}-{\frac {0}{5}}\right)\\&={\frac {2}{15}}.\end{aligned}}$

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