Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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! :-)

share|improve this question
1  
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 2:13
6  
Did you forget to integrate? –  André Nicolas Jun 23 '11 at 2:13
    
@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 2:33
4  
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 2:37
    
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: $\cos^{m}(x),~ \text{where }m \in \mathbb{Z}$. math.stackexchange.com/questions/25730/… –  night owl Jun 23 '11 at 11:14
add comment

3 Answers

up vote 26 down vote accepted

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

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

share|improve this answer
5  
That's got to be one of the coolest integral manipulations I've seen in a long time. –  Nicolas Villanueva Jun 23 '11 at 12:25
    
@Nico: Thanks :-) This is an old trick I learnt a long time back. I liked it so much, i still remember it! –  Aryabhata Jun 23 '11 at 15:50
add comment

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.

share|improve this answer
7  
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 2:41
1  
@Aryabhata: That is really nice, and I think it deserves its own answer. –  Jonas Meyer Jun 23 '11 at 5:25
    
@Jonas: Thanks. Done :-) –  Aryabhata Jun 23 '11 at 5:35
    
@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 9:53
    
@Chris: I agree (and was one of the reason I just commented first here). You might want to check out the answers here: math.stackexchange.com/questions/29980/… –  Aryabhata Jun 23 '11 at 15:45
add comment

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 $2\pi$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.