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

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.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (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: $\color {blue}\int _{0}^{\pi }\sin ^{5}xdx={\frac {16}{15}}$