1. ## integral with cos

Hi everyone,

Can someone give me a hint on how to do this problem, please? I've tried it several different ways, but it doesn't seem to be working.

Find the integral x to the 5thcos(x cubed)dx

u=x to the 3rd
dv=x to the 5th
du=3x2
v=x6/6

(x to the 3rd)(x to the 5th)=(x to the 3rd)(x to the 6th)-int.(x to the 6th/6)3x sq.

but this integral looks worse

Any suggestions?

Thank you

2. Originally Posted by chocolatelover
Hi everyone,

Can someone give me a hint on how to do this problem, please? I've tried it several different ways, but it doesn't seem to be working.

Find the integral x to the 5thcos(x cubed)dx

u=x to the 3rd
dv=x to the 5th
du=3x2
v=x6/6

(x to the 3rd)(x to the 5th)=(x to the 3rd)(x to the 6th)-int.(x to the 6th/6)3x sq.

but this integral looks worse

Any suggestions?

Thank you
do a regular substitution first. Let $u = x^3$

then do by parts on the integral that results

3. I put

t=x3
x2dx=dt/3

int. x to the 5th cos (x3)dx=

int. cosu3dx=

1/3int. cos t3dx

u=cosu3
dv=dx/3
du=-sinu3(3u2)
v=

Can you tell me if this right so far and if dv is right?

Thank you

4. Originally Posted by chocolatelover
I put

t=x3
x2dx=dt/3

int. x to the 5th cos (x3)dx=

int. cosu3dx=

1/3int. cos t3dx

u=cosu3
dv=dx/3
du=-sinu3(3u2)
v=

Can you tell me if this right so far and if dv is right?

Thank you
i have no idea what you just did. let me do the first substitution for you. i leave the integration by parts to you

$\int x^5 \cos \left( x^3 \right) ~dx$

We proceed by substitution
Let $t = x^3$

$\Rightarrow dt = 3x^2~dx$

$\Rightarrow \frac {1}{3}dt = x^2 ~dx$

Now, $\int x^5 \cos \left( x^3 \right) ~dx = \int x^3 \cdot x^2 \cos \left( x^3 \right) ~dx$

$= \frac {1}{3} \int t \cos t~dt$

Now we continue with integration by parts ...

(Hint: obviously we use $u = t$ and $dv = \cos u ~du$)

5. I kind of confused. How do you have t and u in the final interval? In this case I'm using t so that I don't get the u from parts mixed up with the u substitution right? but wouldn't there be two t's instead of the u?

1/3int.tcostdt

Thank you

6. Originally Posted by chocolatelover
I kind of confused. How do you have t and u in the final interval? In this case I'm using t so that I don't get the u from parts mixed up with the u substitution right? but wouldn't there be two t's instead of the u?

1/3int.tcostdt

Thank you
yes, you are correct. i meant to type t, but i'm so used to typing u for substitutions that my hand did it against my will...bad hand! bad! *slap*

7. Would my final answer be tsin-cost?

u=t
dv=costdt
du=1
v=sint

tsin-int.sint=

tsin-cost

Thank you

8. Originally Posted by chocolatelover
Would my final answer be tsin-cost?

u=t
dv=costdt
du=1
v=sint

tsin-int.sint=

tsin-cost

Thank you
it is + cos(t). and you forgot to multiply through by the 1/3

and you also forgot to back substitute to get the x back

we started with x remember, you can't give an answer in terms of t