Quick integral question

1. Jul 12, 2013

Mdhiggenz

1. The problem statement, all variables and given/known data

Hello guys,

I'm studying for the Putnam, and i'm going over problem solving strategies involving symmetry. I got the symmetry portion correct, but their conclution to solving the integral is what confused me. I'm not sure how they got they got from ∫cos^2(x)dx=∫sin^2(x)dx to 1/2∫cos^2(x)+sin^2(x)dx

2. Relevant equations

3. The attempt at a solution

2. Jul 12, 2013

micromass

What integral are you trying to calculate in the first place?

3. Jul 12, 2013

SteamKing

Staff Emeritus
Use the trig identity cos^2 + sin^2 = 1 and re-write the first two integrals.

For example cos^2 = 1 - sin^2 and vice versa.

4. Jul 12, 2013

Mdhiggenz

Still doesn't answer the question to where the 1/2 comes from

5. Jul 12, 2013

SteamKing

Staff Emeritus
This is a homework forum. You've got to show some effort.

6. Jul 12, 2013

micromass

And can you please give us the exact problem statement?

7. Jul 12, 2013

Mdhiggenz

It isn't homework. I even linked you the answer lol.

The problem statement is to compute the integral 0<x<(1/2)∏ ∫cos^2(x)dx in your head.

The example wanted you to use symmetry so if you were able to picture the graph in your head you see that both cos(x) and sin(x) both are symmetric on the above interval.

So what they then do is what is shown in the first picture which I understand.

However what I dont understand is how the manipulate the first picture into the second picture.

Thanks

Higgenz

8. Jul 12, 2013

tiny-tim

Hi Mdhiggenz!

Use the substitution y = π/2 - x in ∫0π/2 sin2x dx …

what do you get?

(and use that if A = B, then A = (A+B)/2)

9. Jul 12, 2013

Saitama

You had $I=\int_{0}^{\pi/2} \cos^2x$. This is equivalent to $I=\int_{0}^{\pi/2} \sin^2x$. Add the two.

This is the most basic stuff taught in integral calculus.

10. Jul 13, 2013

SteamKing

Staff Emeritus
Your OP may not be homework, but it was posted in a homework forum. PF has very explicit rules about what responses are permitted in homework forums. The PF administrators are very diligent about enforcing the rules and pointing out infractions. BTW, it is the folks who respond who acquire these infractions.

Hint: If your question is not homework, please post it in one of the non-homework forums.

11. Jul 13, 2013

micromass

It actually does belong in the homework forums, even if it's not formally homework. This is a textbook-style problem. So it belongs here.

12. Jul 13, 2013

ehild

$$\int_0^{\pi/2}{\cos^2(x)dx}=\int_0^{\pi/2}{\sin^2(x)dx}=A$$

The sum of the two integrals is 2A. But the sum of integrals is the same as the integral of the sum of the integrands. Call that I. You can integral the sum of sin2x+cos2x "in your head" - why? :tongue2:. I=2A. What is A then?

ehild