Integrating Trigonometric Functions: Solving a Challenging Integral Problem

In summary, the given integral can be evaluated by setting x = pi/2 - u and setting the original integral equal to the new integral. This allows for the use of symmetry to solve for the unknown integral, resulting in a solution for the original integral.
  • #1
Slightly
29
0

Homework Statement



[itex]\int\limits_0^\frac{\Pi}{2} \ \frac{x \sin x \cos x}{\sin^4x+\cos^4x} dx[/itex]

Homework Equations



--

The Attempt at a Solution



I first started to try a u-sub, then attempted a change of variables. I tried simplifying the expression, but the closest thing I came to solving this is by using integration by parts as it is the only way to get rid of the x.



[itex]u = x \\

dv = \frac{\sin x \cos x}{\sin^4x+\cos^4x} dx [/itex]

I obtain this:

[itex] - \frac{x}{2} \arctan (\cos (2x)) + \frac{1}{2}\int\limits_0^\frac{\Pi}{2} \arctan (\cos (2x))dx[/itex]
(Of course the first part of the integration by parts is evaluated from 0 to pi/2.
But then, I run into the same problem as before where I don't know how to integral arctan with a cosine inside.
 
Physics news on Phys.org
  • #2
Slightly said:

Homework Statement



[itex]\int\limits_0^\frac{\Pi}{2} \ \frac{x \sin x \cos x}{\sin^4x+\cos^4x} dx[/itex]

Homework Equations



--

The Attempt at a Solution



I first started to try a u-sub, then attempted a change of variables. I tried simplifying the expression, but the closest thing I came to solving this is by using integration by parts as it is the only way to get rid of the x.



[itex]u = x \\

dv = \frac{\sin x \cos x}{\sin^4x+\cos^4x} dx [/itex]

I obtain this:

[itex] - \frac{x}{2} \arctan (\cos (2x)) + \frac{1}{2}\int\limits_0^\frac{\Pi}{2} \arctan (\cos (2x))dx[/itex]
(Of course the first part of the integration by parts is evaluated from 0 to pi/2.
But then, I run into the same problem as before where I don't know how to integral arctan with a cosine inside.

You probably can't integrate that. I think this is more of a trick question using the special limits. Can you show if you change x to pi/2-x and integrate over the same limits, then you should get the same value?
 
  • #3
Dick said:
You probably can't integrate that. I think this is more of a trick question using the special limits. Can you show if you change x to pi/2-x and integrate over the same limits, then you should get the same value?


If I change the x to pi/2-x, would I get the value of the integral?
 
  • #4
Slightly said:
If I change the x to pi/2-x, would I get the value of the integral?

No, you'll get another integral that's equal to the first integral, if you can see why they are equal. Write down the expression saying they are equal. Expand it out a little and see your unknown integral occur twice. Solve for it. Like in integration by parts where you get the same integral back.
 
  • #5
Dick said:
No, you'll get another integral that's equal to the first integral, if you can see why they are equal. Write down the expression saying they are equal. Expand it out a little and see your unknown integral occur twice. Solve for it. Like in integration by parts where you get the same integral back.

But how is this integrating it? When I set x = Pi/2 - μ

so [itex]\int\limits_0^\frac{\Pi}{2} \frac{x \sin x \cos x}{\sin^4x+cos^x} = \int\limits_0^\frac{\Pi}{2} \frac{( \Pi/2 - μ )\sin μ \cos μ}{\sin^4μ+cos^4μ}[/itex]

Set them equal to each other? I think I see what you are saying. Let me work on it.
 
  • #6
Slightly said:
But how is this integrating it? When I set x = Pi/2 - μ

so [itex]\int\limits_0^\frac{\Pi}{2} \frac{x \sin x \cos x}{\sin^4x+cos^x} = \int\limits_0^\frac{\Pi}{2} \frac{( \Pi/2 - μ )\sin μ \cos μ}{\sin^4μ+cos^4μ}[/itex]

Set them equal to each other? I think I see what you are saying. Let me work on it.

Yes, that's exactly what I mean. You've already done the hard part when you were setting up the integration by parts.
 
  • #7
Dick said:
Yes, that's exactly what I mean. You've already done the hard part when you were setting up the integration by parts.

Thank you! I was able to figure it out. All I needed was that push. :P
 

What is an integral?

An integral is a mathematical concept that represents the area under a curve on a graph. It is used to calculate the total value of a function over a specific interval.

Why is "An Interesting Integral" considered interesting?

"An Interesting Integral" is considered interesting because it involves a challenging and unique integration problem that requires creative thinking and advanced mathematical techniques to solve.

Who discovered "An Interesting Integral"?

The integral known as "An Interesting Integral" was first introduced by mathematician Srinivasa Ramanujan, who is famous for his contributions to number theory and mathematical analysis.

What makes "An Interesting Integral" difficult to solve?

"An Interesting Integral" is difficult to solve because it requires a combination of advanced integration techniques and deep understanding of mathematical concepts, making it a challenging problem for even experienced mathematicians.

What are some real-world applications of integrals?

Integrals have a wide range of real-world applications, including calculating areas and volumes, calculating work and energy in physics, and determining probabilities in statistics. They are also used in engineering, economics, and many other fields.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
269
  • Calculus and Beyond Homework Help
Replies
15
Views
744
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
619
  • Calculus and Beyond Homework Help
Replies
1
Views
201
  • Calculus and Beyond Homework Help
Replies
1
Views
449
  • Calculus and Beyond Homework Help
Replies
8
Views
826
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
637
  • Calculus and Beyond Homework Help
Replies
3
Views
488
Back
Top