MATH101 April 2013
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Question 07 (a)

FullSolution 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!

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Solution

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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,
 $\displaystyle \cos 2x=\cos ^{2}x\sin ^{2}x=2\cos ^{2}x1$
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 cos^{4} 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:
 $\displaystyle {\begin{aligned}\int _{0}^{\pi /4}\cos ^{4}\theta \ d\theta &={\frac {1}{8}}\int _{0}^{\pi /4}[\cos 4\theta +4\cos 2\theta +3]\ d\theta \\&={\frac {1}{8}}\left[{\frac {1}{4}}\sin 4\theta +2\sin 2\theta +3\theta \right]{\big }_{0}^{\pi /4}\\&={\frac {1}{8}}\left[{\frac {1}{4}}\sin \pi +2\sin {\frac {\pi }{2}}+3{\frac {\pi }{4}}{\frac {1}{4}}\sin 02\sin 03\cdot 0\right]\\&={\frac {1}{8}}\left[2+{\frac {3\pi }{4}}\right]\\&={\frac {1}{4}}+{\frac {3\pi }{32}}\\&=(8+3\pi )/32\end{aligned}}$
as required.

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