0
$\begingroup$

How would one go about solving the following problem $$\int x\sin(x^2) + 2x^3\;dx?$$ Obviously, I'm stuck with $\sin(x^2)$, and from what I've looked into, there appears to be no elementary antiderivative for this...

  • 1
    do you mean sin(x^2)*2x^3?2012-10-17
  • 1
    You said it yourself, there's no elementary anti-derivative. So what exactly is it that you want?2012-10-17
  • 0
    @EuYu, Well, it's a homework question. I'm just stuck as to how to go forward with this problem.2012-10-17
  • 0
    Are you sure it's given correctly?2012-10-17
  • 0
    I made an edit.2012-10-17

3 Answers 3

1

Make the substitution $u = x^2$, then we have $du/2x = dx$

$$\int x\sin(x^2) + 2x^3 dx = \int \frac{x\sin(u) + 2xu}{2x}du$$

and now we have a simple integral in $u$,

$$\int \frac{\sin(u)}{2} + u \ du$$

which I shall leave you to solve!

0

In case what you really want is $\int2x^3\sin(x^2)\,dx$, let $u=x^2$, then it's an easy integration by parts.

EDIT: with the revised question, all you need is $u=x^2$.

  • 0
    So I guess, even in this case I could do $\int x sin(x^2) dx$ using integration by parts?2012-10-17
  • 0
    No, just $u=x^2$ will do.2012-10-17
0

Suppose you know

$$\int f(x)\,dx= F(x)\Longrightarrow \int f(g(x))g'(x)\,dx=F(g(x))$$ .

Well, now we use this and linearity of indefinite integral:

$$\int (x\sin x^2+2x^3)\,dx=\frac{1}{2}\int \sin x^2\,(2x\,dx)+2\int x^3\,dx=$$

$$=-\frac{1}{2}\cos x^2+...\text{etc}$$

  • 0
    I'm sorry, you're taking the route that I myself found myself taking when I attempted this problem. I like Raymond Cheng's take on the problem--and it appears to be the correct approach.2012-10-18
  • 0
    I'm sorry but I don't care: I can't guess what route you were taking and it is your choice what answer you like the best. Anyway, my answer's correct and I like it better as it doesn't involve substitution. There is no "correct/incorrect" in these two answers (Raymond's and mine), as they both are correct. It's just a matter of taste.2012-10-18