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

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MATH101 April 2017

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Question 05 (a)
Evaluate $\int _{0}^{\pi }\sin ^{5}xdx$ . A calculator-ready answer is sufficient.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Consider the substitution $u=\cos(x)$ and use an appropriate trigonometric identity.

Hint 2
(For the alternative solution) Use integration by parts, using $f^{\prime }(x)=\sin x$ . Use the trig identity $\cos ^{2}x=1-\sin ^{2}x$ to have an expression with only powers of the sine function move all the terms with the highest power to the same side. repeat the process until you get an easy integral you can solve.

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

Solution 2
(Alternative solution) Using integration by parts $\int f^{\prime }g=fg-\int fg^{\prime }$ , we pick $f^{\prime }(x)=\sin x,\,g(x)=\sin ^{4}(x)$ , which yields $\int _{0}^{\pi }\sin ^{5}xdx=-\cos x\sin ^{4}(x)\|_{0}^{\pi }-\int _{0}^{\pi }-\cos x4\sin ^{3}x\cos xdx$ $=4\int _{0}^{\pi }\cos ^{2}x\sin ^{3}xdx$ Using the trig identity $\cos ^{2}x=1-\sin ^{2}x$ , we have $\int _{0}^{\pi }\sin ^{5}xdx=4\int _{0}^{\pi }\sin ^{3}x-\sin ^{5}xdx$ Thus, $\int _{0}^{\pi }\sin ^{5}xdx={\frac {4}{5}}\int _{0}^{\pi }\sin ^{3}xdx$ . We now compute $\int _{0}^{\pi }\sin ^{3}xdx$ using integration by parts with $f^{\prime }(x)=\sin x,\,g(x)=\sin ^{2}x$ , and so $\int _{0}^{\pi }\sin ^{3}xdx=-\cos x\sin ^{2}x-\int _{0}^{\pi }-2\cos ^{2}x\sin xdx$ . Using the same trigonometric identity we have $\int _{0}^{\pi }\sin ^{3}xdx={\frac {2}{3}}\int _{0}^{\pi }\sin xdx$ $=-{\frac {2}{3}}\cos x\|_{0}^{\pi }={\frac {4}{3}}$ . Using $\int _{0}^{\pi }\sin ^{5}xdx={\frac {4}{5}}\int _{0}^{\pi }\sin ^{3}xdx$ we get Answer: $\color {blue}\int _{0}^{\pi }\sin ^{5}xdx={\frac {16}{15}}$