Integration by parts, I on this problem.

[afcwestwarrior]

Homework Statement

∫ (theta)^3 *cos(theta)^2

Homework Equations

integration by parts ∫u dv= uv- ∫v du

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

 

[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.

 

[afcwestwarrior]

ok so u=cos(theta)^2

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

 

[rock.freak667]

wait...was your question


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

OR

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

 

[afcwestwarrior]

my question was the 1st one you put

 

[rock.freak667]

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

 

[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

 

[afcwestwarrior]

v= (1/4) theta^4

 

[rock.freak667]

ok so u=cos(theta)^2

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

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}

Check back your du.

 

[afcwestwarrior]

oh i did

 

[afcwestwarrior]

i forgot to put sin

 

[afcwestwarrior]

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

 

[rock.freak667]

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

 

[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

 

[afcwestwarrior]

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

 

[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.

 

[afcwestwarrior]

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

 

[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)).

 

[afcwestwarrior]

oh ok