# Integration by parts question

1. Jan 3, 2015

### mrg

The problem statement, all variables and given/known data
$$\int x^{3}cos(x^{2})dx$$

The attempt at a solution
OK, so I am aware that there is a way in which to do this problem where you do a substitution (let $$u=x^{2}$$ to do a substitution before you integrate by parts), and I was able to get the answer right using this method. The thing is, I tried it a different way first, and after triple checking it, I feel like I should have gotten it right. It doesn't matter what I choose to let u and dv equal, right? I should get the right answer no matter what I choose (assuming I did things correctly, of course).
Here's my work:

$$u=cos(x^{2})$$
$$du=-2xsin(x^{2})$$
$$v=\frac{x^{4}}{4}$$
$$dv=x^{3}dx$$

$$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}\int x*sin(x^{2})dx$$

$$u_2=sin(x^{2})$$
$$du_2=2xcos(x)$$
$$v_2=\frac{x^{2}}{2}$$
$$dv_2=x$$

$$\frac{x^{4}cos(x^{2})}{4}+\frac{1}{2}(\frac{x^{2}sin(x^{2})}{2}-\int x^{3}cos(x^{2})dx)$$

Since this last integral is the same as the one we started with, we can now say that
$$I=A+B-\frac{1}{2}I$$
Hence
$$\frac{3}{2}I=A+B$$
Multiplying both sides by two over three...
$$\frac{x^{4}cos(x^{2})}{6}+\frac{x^{2}sin(x^{2})}{6}$$

Thanks for your time and any help you can offer me.

2. Jan 3, 2015

### SammyS

Staff Emeritus
So, are you asking why your result isn't correct?

$\displaystyle v\,du = (-1/2)x^5\sin(x^2)\,dx\ \text{ not } (-1/2)x\sin(x^2)\,dx\ .$

3. Jan 3, 2015

### mrg

I knew it was something stupid. No idea where I got x from. Thanks Sammy.

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