How can substitution make integration by parts easier?

[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

 

[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?

 

[Mark44]

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


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


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

the last integral

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.

 

[nameVoid]

right..thank you.

 

[NastyAccident]

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


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


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

the last integral

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:

Spoiler

u = p dv = cos(p) dp
du = dp v = sin(p)

NastyAccident