Science:Math Exam Resources/Courses/MATH101/April 2013/Question 07 (a)

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MATH101 April 2013

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Question 07 (a)
Full-Solution Problem. Justify your answers and show all your work. Simplification of numerical answers is required. Show that $\displaystyle \int _{0}^{\pi /4}\cos ^{4}\theta \ d\theta =(8+3\pi )/32$

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
Think about the double angle formulas.

Hint 2
You have to use double angles twice!

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

The question asks us to confirm that $\displaystyle \int _{0}^{\pi /4}\cos ^{4}\theta \ d\theta =(8+3\pi )/32$ So we will calculate the integral on the left hand side, and then compare our answer to the given right hand side. The key identity is the double angle formula for cosine, $\displaystyle \cos 2x=\cos ^{2}x-\sin ^{2}x=2\cos ^{2}x-1$ where the second equality came from $\displaystyle \sin ^{2}x+\cos ^{2}x=1$. This gives $\displaystyle \cos ^{2}x={\frac {1}{2}}(\cos 2x+1)$ and hence $\displaystyle \cos ^{4}x=\left({\frac {1}{2}}(\cos 2x+1)\right)^{2}$ Then with $\displaystyle x=\theta$, $\displaystyle {\begin{aligned}\int _{0}^{\pi /4}\cos ^{4}\theta \ d\theta &=\int _{0}^{\pi /4}\left({\frac {1}{2}}(\cos 2\theta +1)\right)^{2}\ d\theta \\&={\frac {1}{4}}\int _{0}^{\pi /4}(\cos ^{2}2\theta +2\cos 2\theta +1)\ d\theta \end{aligned}}$ Note that this is not a substitution in the integral. All we do is rewrite cos4 θ using the double angle formula. We notice that we have a term that looks like $\displaystyle {}\cos ^{2}{\text{(something)}}$ again. The identity above holds for any argument and so we could write $\displaystyle \cos ^{2}\clubsuit ={\frac {1}{2}}(\cos 2\clubsuit +1).$ We have $\displaystyle {}\cos ^{2}(2\theta )$ and so we can replace $\clubsuit$ with $\displaystyle 2\theta$ $\displaystyle {\begin{aligned}\int _{0}^{\pi /4}\cos ^{4}\theta \ d\theta &={\frac {1}{4}}\int _{0}^{\pi /4}(\cos ^{2}2\theta +2\cos 2\theta +1)\ d\theta \\&={\frac {1}{4}}\int _{0}^{\pi /4}\left[{\frac {1}{2}}(\cos(2(2\theta ))+1)+2\cos 2\theta +1\right]\ d\theta \end{aligned}}$ To make things cleaner, we'll factor out $\displaystyle 1/2$ from every summand in the integral: $\displaystyle {\begin{aligned}\int _{0}^{\pi /4}\cos ^{4}\theta \ d\theta &={\frac {1}{4}}\int _{0}^{\pi /4}\left[{\frac {1}{2}}(\cos 4\theta +1)+2\cos 2\theta +1\right]\ d\theta \\&={\frac {1}{8}}\int _{0}^{\pi /4}[(\cos 4\theta +1)+4\cos 2\theta +2]\ d\theta \\&={\frac {1}{8}}\int _{0}^{\pi /4}[\cos 4\theta +4\cos 2\theta +3]\ d\theta \end{aligned}}$ Now, we can integrate each term, remembering the scaling factor: $\displaystyle {\begin{aligned}\int _{0}^{\pi /4}\cos ^{4}\theta \ d\theta &={\frac {1}{8}}\int _{0}^{\pi /4}[\cos 4\theta +4\cos 2\theta +3]\ d\theta \\&={\frac {1}{8}}\left[{\frac {1}{4}}\sin 4\theta +2\sin 2\theta +3\theta \right]{\big |}_{0}^{\pi /4}\\&={\frac {1}{8}}\left[{\frac {1}{4}}\sin \pi +2\sin {\frac {\pi }{2}}+3{\frac {\pi }{4}}-{\frac {1}{4}}\sin 0-2\sin 0-3\cdot 0\right]\\&={\frac {1}{8}}\left[2+{\frac {3\pi }{4}}\right]\\&={\frac {1}{4}}+{\frac {3\pi }{32}}\\&=(8+3\pi )/32\end{aligned}}$ as required.

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