Symmetry in Problem Solving: Understanding Quick Integrals"

  • Thread starter Mdhiggenz
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In summary, the conversation was about solving an integral using symmetry. The question was how to manipulate ∫cos^2(x)dx=∫sin^2(x)dx to 1/2∫cos^2(x)+sin^2(x)dx. The solution involved using the trig identity cos^2 + sin^2 = 1 and the property that the sum of integrals is the integral of the sum of the integrands. The final answer was 1/2∫cos^2(x)+sin^2(x)dx, which can be solved mentally.
  • #1
Mdhiggenz
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Homework Statement



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



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Homework Equations





The Attempt at a Solution

 
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  • #2
What integral are you trying to calculate in the first place?
 
  • #3
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
Still doesn't answer the question to where the 1/2 comes from
 
  • #5
Mdhiggenz said:
Still doesn't answer the question to where the 1/2 comes from

This is a homework forum. You've got to show some effort.
 
  • #6
And can you please give us the exact problem statement?
 
  • #7
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 don't understand is how the manipulate the first picture into the second picture.

Thanks

Higgenz
 
  • #8
Hi Mdhiggenz! :smile:

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

what do you get? :wink:

(and use that if A = B, then A = (A+B)/2)
 
  • #9
Mdhiggenz said:
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 don't understand is how the manipulate the first picture into the second picture.

Thanks

Higgenz

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
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
SteamKing said:
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.

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
Mdhiggenz said:
However what I don't understand is how the manipulate the first picture into the second picture. Higgenz

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

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? :-p. I=2A. What is A then?

ehild
 

FAQ: Symmetry in Problem Solving: Understanding Quick Integrals"

What is a quick integral question?

A quick integral question is a type of problem in mathematics that involves finding the area under a curve or the volume of a solid. It requires knowledge of integration techniques and often involves solving for an unknown variable.

Why are integrals important?

Integrals are important because they allow us to calculate the area, volume, and other important properties of complex shapes and curves. They also have many practical applications in fields such as physics, engineering, and economics.

How do I solve a quick integral question?

To solve a quick integral question, you first need to identify the type of integral (definite or indefinite) and the appropriate integration technique (substitution, integration by parts, etc.). Then, you can follow the steps of the chosen technique to solve for the unknown variable.

What are some tips for solving quick integral questions?

Some tips for solving quick integral questions include: identifying the type of integral, choosing the appropriate integration technique, carefully evaluating the limits of integration, and checking your answer for correctness. It is also helpful to practice and become familiar with different integration techniques.

Can I use a calculator to solve quick integral questions?

While a calculator can be helpful for evaluating integrals with complex functions or large numbers, it is important to have a strong understanding of integration techniques and how to solve integrals by hand. It is recommended to use a calculator as a tool to check your work, rather than relying on it to solve the entire problem.

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