8

So when looking at 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! :-)

InBetween
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    You need to find the antiderivative of $cos^2(x)$ first. How would you compute $\int_0^\pi cos(x)dx$? – Adam Saltz Jun 23 '11 at 02:13
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    Did you forget to integrate? – André Nicolas Jun 23 '11 at 02:13
  • @IAmBrianDawkins, well the antiderivative of cos(x) would just be sin(x) right? So 3sin^3(x) for be the antiderivative? – InBetween Jun 23 '11 at 02:33
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    You know how to differentiate. Check whether the derivative of your candidate for the answer really is $\cos^2 x$. Then look at the integration hint provided by yunone. – André Nicolas Jun 23 '11 at 02:37
  • 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#33438 – night owl Jun 23 '11 at 11:14

3 Answers3

31

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

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

Aryabhata
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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.

yunone
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    Another way is: $\int_{0}^{\pi} \cos^2 x = 2 \int_{0}^{\pi/2} \cos^2 x = 2 \int_{0}^{\pi/2} \sin^2 x$. Adding the last two gives the answer. – Aryabhata Jun 23 '11 at 02:41
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    @Aryabhata: That is really nice, and I think it deserves its own answer. – Jonas Meyer Jun 23 '11 at 05:25
  • @Jonas: Thanks. Done :-) – Aryabhata Jun 23 '11 at 05:35
  • @Aryabhata that's a nice trick, though I think for pedagogical purposes I prefer yunone's answer (+1 to both, though) – Chris Taylor Jun 23 '11 at 09:53
  • @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-textdx – Aryabhata Jun 23 '11 at 15:45
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 $2\pi$.

Américo Tavares
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