• Support PF! Buy your school textbooks, materials and every day products Here!

Definite integration

  • Thread starter Saitama
  • Start date
  • #1
3,812
92

Homework Statement


Solve:
[tex]\int_{0}^{4/ \pi} 3x^2 \cdot sin(\frac{1}{x})-x \cdot cos(\frac{1}{x})dx[/tex]

Homework Equations





The Attempt at a Solution


I am stuck on this question because i really have no idea what should i be doing here with sin(1/x) and cos(1/x). I do not know how to integrate them. Even if i try to integrate by parts, i am stuck with these sin(1/x) and cos(1/x).

Any help is appreciated.
Thanks!
 

Answers and Replies

  • #2
296
0
First, notice that you can split the integral into two different integrals. I will do one of them.

[tex]\int_{0}^{4/\pi}x\cos\left(\frac{1}{x}\right)\,dx=\int_{\pi/4}^{\infty}\frac{\cos(t)}{t^3}\,dt[/tex]

The expansion of this integral involves the cosine integral, the case with sine is not very different. This makes me think that the integral has no elementary expansion, but I'm not sure. There might be a way to do it without splitting the integral.
 
  • #3
3,812
92
The expansion of this integral involves the cosine integral, the case with sine is not very different.
I have no idea what you meant by the "cosine integral".
If i rewrite the expression as you said:
[tex]\int_{\pi/4}^{\infty} \frac{3\sin(t)}{t^4}-\frac{\cos(t)}{t^3}\,dt[/tex]
What should i be doing next?
 
  • #4
6,054
390
The indefinite integral has no representation via elementary functions. The definite integral has to be computed in some other way, e.g., via the residue theorem of complex analysis.
 
  • #5
3,812
92
The indefinite integral has no representation via elementary functions. The definite integral has to be computed in some other way, e.g., via the residue theorem of complex analysis.
Well, if you plug in the indefinite integral in wolfram alpha, the answer it shows is x3sin(1/x).
 
  • #6
256
0
I beg to differ ,
This question can be tackled , by using integration by parts ,
Let me give you a hint Pranav arora
If you integrate 3sin(t)/t^4 , using integration by parts.
Then one of two terms would be cos(t)/t^3 , which gets cancelled . and you are left with a simple integral which can be solved very easily :)
 
  • #7
6,054
390
Indeed. The addends cannot be integrated separately, but together the nicely cancel out the nasty part.
 
  • #8
3,812
92
I beg to differ ,
This question can be tackled , by using integration by parts ,
Let me give you a hint Pranav arora
If you integrate 3sin(t)/t^4 , using integration by parts.
Then one of two terms would be cos(t)/t^3 , which gets cancelled . and you are left with a simple integral which can be solved very easily :)
Excellent, thanks kushan! That really worked! :smile:
 
  • #9
256
0
Your welcome :cool:
 
  • #10
ehild
Homework Helper
15,477
1,854
Nice trick, Kushan!

Pranav, you should think of a trick if you are given such a problem.

Try to consider the integrand as the derivative of a product (fg)=f'g+fg'.

3x^2 is the derivative of x^3. cos() is the derivative of sin(), so find the derivative of sin(1/x)... It is -cos(1/x)/x^2. f'g+fg'=3x^2sin(1/x)-x^2cos(1/x)1/x^2= 3x^2sin(1/x)-xcos(1/x)

ehild
 
  • #11
3,812
92
3x^2 is the derivative of x^3. cos() is the derivative of sin(), so find the derivative of sin(1/x)...
Man, your trick made the question lot easier than before, thank you very much ehild! I will always try to use this wherever possible. :smile:

As for the derivative of sin(1/x), it is -1/(x2)*cos(1/x).
So according to the trick, the answer is x3sin(1/x). If i differentiate this, i get the same equation as mentioned in the question.
 
  • #12
ehild
Homework Helper
15,477
1,854
Right. Suck tricks are very useful in school, but hardly ever applicable in real life... Life is ugly.:tongue2:

ehild
 
  • #13
256
0
3x^2 is the derivative of x^3. cos() is the derivative of sin(), so find the derivative of sin(1/x)... It is -cos(1/x)/x^2. f'g+fg'=3x^2sin(1/x)-x^2cos(1/x)1/x^2= 3x^2sin(1/x)-xcos(1/x)

ehild
Yes this is really helpful , and these kind of question are very common here , they look impossible though at first look .
 

Related Threads on Definite integration

Replies
7
Views
3K
  • Last Post
Replies
7
Views
654
Replies
5
Views
599
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
1
Views
745
Top