# Integration by parts

1. Oct 7, 2009

### nameVoid

$$\int x^3cos(x^2)dx$$

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

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

Last edited: Oct 7, 2009
2. Oct 7, 2009

### tiny-tim

Hi nameVoid!

(i think that the minus is wrong, and where did the 3/2 come from? but anyway …)

Where did the last integral come from? What did you think you were differentiating?

3. Oct 7, 2009

### Staff: Mentor

In your integration by parts, you don't show u and dv, etc., but you seem to be mostly on the right track.

For the first integral, u = x2 and dv = x cos(x2) dx. So du = 2xdx and v = -(1/2)sin(x2).
So
$$\int x^3cos(x^2)dx$$
$$=~-(1/2)x^2sin(x^2) + \int x sin(x^2)dx$$
So far, the only difference between your work and mine is the multiplier in front of the last integral. This one can be done with an ordinary substitution (u = x2), instead of the integration by parts that it looks like you tried.

4. Oct 7, 2009

### nameVoid

right..thank you.

5. Oct 7, 2009

### NastyAccident

Just a suggestion, make life a little easier on yourself by doing a substitution before int by parts:

$$\int x^3cos(x^2)dx$$

$$let p = x^2$$

$$dp = 2xdx$$

$$\frac{1}{2}*\int 2x^3cos(x^2)dx$$

$$\frac{1}{2}*\int pcos(p)dp$$

From there, the integration by parts is soo much simpler.

Here's the variable breakdown for integration by parts:
u = p dv = cos(p) dp
du = dp v = sin(p)

Sincerely,

NastyAccident