Question

how do you integrate x^2cos(x^3)dx

Answer 1

Let z=x²
so dz=2xdx
The integral becomes:
∫1/2 z cosz dz

Using by parts
u=z, dv=cosz
du=1,v=sinz
So ∫1/2 z cosz dz
=1/2 [zsinz - ∫sinzdz]
=1/2 [zsinz+cosz] + c
=1/2(x²sinx² + cosx²} + c

Answer 2

do this help

Answer 3

yes im honestly just stuck on what to do with cos(x^3) because i know what to do with the other stuff (meaning integration by parts)

Answer 4

and im kinda lost on what z and dz is because i use u,du dv,v

Answer 5

yes

Answer 6

dont use that one no more

Answer 7

I hope you didn't copy this "solution" from somewhere online! @TheRealMeeeee

Answer 8

no i didnt not this time

Answer 9

Hard to believe. You typed so much in a few moments... can you even EXPLAIN your solution?

Answer 10

well my computer is mesed up it didnt show how long i was typing till the end

Answer 11

Yeah right.

Answer 12

ok watever

Answer 13

my real question is how do you integrate cos(x^3) because i know that is what you use for dv

Answer 14

and i know you got that from yahoo i saw it before

Answer 15

ok watever

Answer 16

ok

Answer 17

so how do i integrate cos(x^3)

Answer 18

because i know its by parts

Answer 19

just set u = x^3 , then du = 3x^2

Answer 20

instead do this:

u = x^3
du = 3x^2 dx

\[\int\limits_{}^{}x^2 \cos(x^3) dx = \int\limits_{}^{}\frac{ 1 }{ 3 }\cos(u)du = \frac{ sinu }{ 3 } + C = \frac{ \sin(x^3) }{ 3 } + C\]

Answer 21

thanks you!

Answer 22

@TheRealMeeeee youre answer is wrong. read the problem carefully

Answer 23

See the way its done? @TheRealMeeeee

Answer 24

Euler has the correct answer, a simple sub is all you need.

Answer 25

yes i see

Answer 26

thank you i thought i had to do integration by parts