1. Integration with parts repeatedly

$\int x^5cos(x^3)dx$

dv = cos(x^3)
u= x^5
v = sin x^3 / 3
du = 5x^4

With those settings I can see that I'm going to have to integrate by parts several times, and by several I mean more than 4 times. Is there a faster way to solve this problem that I don't realize?

2. Originally Posted by Archduke01
$\int x^5cos(x^3)dx$

dv = cos(x^3)
u= x^5
v = sin x^3 / 3
du = 5x^4

With those settings I can see that I'm going to have to integrate by parts several times, and by several I mean more than 4 times. Is there a faster way to solve this problem that I don't realize?
Your $v$ is wrong.
You can not find $v$ in terms of the elementary functions.
You need the gamma function. Which is ,surely, you did not cover it yet.

If I were you, I will rewrite $x^5$ as $x^3 \, x^2$ then I will use the substitution $t=x^3$ then I will integrate it by parts

3. Originally Posted by Archduke01
$\int x^5cos(x^3)dx$

dv = cos(x^3)
u= x^5
v = sin x^3 / 3
du = 5x^4

With those settings I can see that I'm going to have to integrate by parts several times, and by several I mean more than 4 times. Is there a faster way to solve this problem that I don't realize?
If you use $u=x^3$

then $\frac{du}{dx}=3x^2\ \Rightarrow\ du=3x^2dx$

$\frac{1}{3}udu=x^5dx$

the integral becomes

$\frac{1}{3}\int{ucos(u)}du$

4. Originally Posted by Miss
Your $v$ is wrong.
You can not find $v$ in terms of the elementary functions
But, then what is my v if my dv is cos (x^3)?

5. Originally Posted by Archduke01
But, then what is my v if my dv is cos (x^3)?
I told you can not find v in terms of the elementary functions.

6. $\frac{1}{3}\int{ucosu}du$

$dv=cosudu\ \Rightarrow\ v=sinu$

$\frac{1}{3}\int{ucosu}du=\frac{1}{3}\left(usinu-\int{sinu}du\right)$

$=\frac{1}{3}\left(x^3sin\left(x^3\right)--cos\left(x^3\right)\right)+C$

$=\frac{1}{3}\left(x^3sin\left(x^3\right)+cos\left( x^3\right)\right)+C$