Definite Integration: Solving Homework Problem

[Saitama]

Homework Statement

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

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!

 

[Millennial]

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

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

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.

 

[Saitama]

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:
$$\int_{\pi/4}^{\infty} \frac{3\sin(t)}{t^4}-\frac{\cos(t)}{t^3}\,dt$$
What should i be doing next?

 

[voko]

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.

 

[Saitama]

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).

 

[kushan]

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 canceled . and you are left with a simple integral which can be solved very easily :)

 

[voko]

Indeed. The addends cannot be integrated separately, but together the nicely cancel out the nasty part.

 

[Saitama]

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 canceled . and you are left with a simple integral which can be solved very easily :)

Excellent, thanks kushan! That really worked!

 

[kushan]

Your welcome

 

[ehild]

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

 

[Saitama]

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.

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.

 

[ehild]

Right. Suck tricks are very useful in school, but hardly ever applicable in real life... Life is ugly.

ehild

 

[kushan]

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 .