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So when looking on the question:

$\int_{0}^{\pi} \cos^2 x \ \text{d}x$

I would just subtract $\cos^2(0)$ from $\cos^2(\pi)$, but doing so would get me 1 - 1 = 0. When the answer is $\pi/2$. Where did I go wrong? What am I missing? Thanks so much for all your help! :-)

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    You can take a look at the link that will be provided to see how to evaluate cosine to any power of integers as such: $\cos^{m}(x),~ \text{where }m \in \mathbb{Z}$. http://math.stackexchange.com/questions/25730/the-integral-int-cos32x-mathrm-dx/33438#334382011-06-23

3 Answers 3

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We have that

$I = \int_{0}^{\pi} \cos^2 x \ dx= 2 \int_{0}^{\pi/2} \cos^2 x \ dx = 2 \int_{0}^{\pi/2} \cos^2 (\pi/2 - x) \ dx = 2 \int_{0}^{\pi/2} \sin^2 x \ dx$

and thus

$I = \int_{0}^{\pi/2} (\cos^2 x +\sin^2 x)\ dx = \pi/2$

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    @Nico: Thanks :-) This is an old trick I learnt a long time back. I liked it so much, i still remember it!2011-06-23
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You need to integrate the integrand $\cos^2(x)$ first. The identity $\displaystyle\cos^2(x)=\frac{1+\cos(2x)}{2}$ is of use here.

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    @Chris: I agree (and was one of the reason I just commented first here). You might want to check out the answers here: http://math.stackexchange.com/questions/29980/evaluating-int-p-sin-x-cos-x-textdx2011-06-23
5

From the addition identity:

$\cos (a+b)=\cos a\cdot \cos b-\sin a\cdot \sin b,$

we get (setting $a=b$)

$\cos (2a)=\cos ^{2}a-\sin ^{2}a.$

Applying the Pythagorean trigonometric identity $\cos^2a+\sin^2a=1$, in the form $\sin^2a=1-\cos^2a,$

yields

$\cos (2a)=\cos ^{2}a-\sin ^{2}a=\cos ^{2}a-1+\cos^2a=2\cos ^{2}a-1,$

or, equivalently

$\cos ^{2}a=\dfrac{1+\cos (2a)}{2}.$

Setting $x=a$ results in

$\cos ^{2}(x)=\dfrac{1+\cos (2x)}{2}=\dfrac{1}{2}+\dfrac{\cos (2x)}{2}.$

Then

$\int_{0}^{\pi} \cos^2 x \ \text{d}x=\int_{0}^{\pi}\dfrac{1}{2}+\dfrac{\cos (2x)}{2} \ \text{d}x=\dfrac{1}{2}\pi+\dfrac{1}{2}\int_{0}^{\pi}\cos (2x)\ \text{d}x=\dfrac{1}{2}\pi+\dfrac{1}{4}\int_{0}^{2\pi }\cos t\;\mathrm{d}t.$

I leave to you the evaluation of $\displaystyle\int_{0}^{2\pi}\cos t\ \text{d}t$. Remember that you have to find the antiderivative of $\cos t$, or just observe that the period of $\cos t$ is equal to $2\pi$.