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...
How to do this integral?
0
$\begingroup$
calculus
integration
-
0I made an edit. – 2012-10-17
3 Answers
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 $u=x^2$.
-
0No, 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:
$\int (x\sin x^2+2x^3)\,dx=\frac{1}{2}\int \sin x^2\,(2x\,dx)+2\int x^3\,dx=$
$=-\frac{1}{2}\cos x^2+...\text{etc}$
-
0I'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