Integration by parts, I need help on this problem.

1. Aug 23, 2008

afcwestwarrior

1. The problem statement, all variables and given/known data
∫ (theta)^3 *cos(theta)^2

2. Relevant equations
integration by parts ∫u dv= uv- ∫v du

3. The attempt at a solution
u=theta^3 dv=cos(theta)^2
du=3theta^2 v=sin(theta)^2

here's the problem do i use cos(theta)^2 equal to u, because i'm not sure if the antiderivative i got is right

this is where i'm stuck

2. Aug 23, 2008

rock.freak667

If you wrote the problem correctly, then you should put $u=cos(\theta^2)$ since the anti-derivative of it can't be expressed in a way which will help you.

3. Aug 23, 2008

afcwestwarrior

ok so u=cos(theta)^2

du=2(theta) *sin(theta)^2

Last edited: Aug 23, 2008
4. Aug 23, 2008

rock.freak667

$$\int \theta^3 cos(\theta^2) d\theta$$

OR

$$\int \theta^3 (cos\theta)^2 d\theta$$

5. Aug 23, 2008

afcwestwarrior

my question was the 1st one you put

6. Aug 23, 2008

rock.freak667

well in that case. $dv=\theta^3 d\theta$. v=?. Then put it in the formula.

7. Aug 23, 2008

afcwestwarrior

ok i think i'm stuck again now that i have my equation it's
(1/4 cos) (theta)^2 * (theta)^4 - 1/2 ∫theta^4 *(theta)sin(theta)^2
=(1/4) cos (theta)^4 * (theta)^4 -1/2 ∫ (theta)^4 *(theta) sin(theta)^2

i edited it

Last edited: Aug 23, 2008
8. Aug 23, 2008

afcwestwarrior

v= (1/4) theta^4

9. Aug 23, 2008

rock.freak667

$$u=cos(\theta^2)$$

Let $t=\theta^2 \Rightarrow \frac{dt}{d\theta}=2\theta$

$$u=cos(t) \Rightarrow \frac{du}{dt}=-sin(t)$$

$$\frac{du}{d\theta}=\frac{du}{dt}*\frac{du}{d\theta}$$

Last edited: Aug 23, 2008
10. Aug 23, 2008

afcwestwarrior

oh i did

11. Aug 23, 2008

afcwestwarrior

i forgot to put sin

12. Aug 23, 2008

afcwestwarrior

ok so u=cos (theta)^2
du= - 2(theta) sin theta^2

13. Aug 23, 2008

rock.freak667

You might need to another integration by parts and see if it simplifies.

14. Aug 23, 2008

afcwestwarrior

=(1/4) cos (theta)^4 * (theta)^4 -1/2 ∫ (theta)^5 * sin(theta)^2

so i take this part ∫ (theta)^4 * sin(theta)^2
u=sin (theta)^2
du= 2 theta * cos (theta)^2
dv=(theta)^5
v=(1/6) (theta^6)

it looks like i'll keep on integrating

15. Aug 23, 2008

afcwestwarrior

sin (theta)^2 *1/6) (theta^6)-∫ 1/6) (theta^6*2 theta * cos (theta)^2

16. Aug 23, 2008

bigevil

afcwestwarrior, that's not necessary.

Notice that you can split the (I'm going to make x = theta) x^3 cos (x^2) into 0.5(x^2)(2x cos (x^2)). You solve the problem by integrating the 2x cos (x^2) term and differentiating the 0.5x^2 term.

17. Aug 23, 2008

afcwestwarrior

how'd u make it into 0.5(x^2)(2x cos (x^2)).

18. Aug 23, 2008

bigevil

Break the x^3 term into x^2 and x.

When you differentiate x^2 you get 2x, so what you need to do is to create a 2x term. So the integral breaks down into 0.5x^2 (2x cos (x^2)).

19. Aug 23, 2008

oh ok