Question

whats the integral of sin x^(3) 3x^(2)?

Answer 1

1/36 (-81 (x^2-2) cos(x)+(9 x^2-2) cos(3 x)-6 x (sin(3 x)-27 sin(x)))+C

Answer 2

i know i have to use u du = -cos u +c, but dont know how to do it step by step :/

Answer 3

LEt x^3 = u

see if it works ?

Answer 4

-cos(x)^(3) + C is that correct?

Answer 5

but what happened with du (which equals to 3x^2 ) ?

Answer 6

make the function as 3x^2* sinx^3

Answer 7

factor out 3

Answer 8

do u know trig identities of sin^2 X??

Answer 9

separate sin^3x into Sin^2x Sinx

Answer 10

its now become 3Integral X^2(sinX)(1/2(1-cos2x)dx

Answer 11

convert that into 3/2 int x^2(sinx)(1-cos2x)dx

Answer 12

Use the trigonometric identity sin(a) cos(b)
3/2 integral x^2 sin(x)dx - 3/4
integral x^2 (sin(3 x)-sin(x)) dx

Answer 13

its become 3/2 integral x^2 sin(x) dx - 3/4integral (x^2 sin(3 x)-x^2 sin(x)) dx

Answer 14

Integrate the sum term by term and factor out constants:
= 3/4 integral x^2 sin(x) dx-3/4 integral x^2 sin(3 x) dx+3/2 integral x^2 sin(x) dx

Answer 15

im just gonna write it out what i do next.. just copy in out then u'll see

Answer 16

1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/4 integral x^2 sin(x) dx-1/2 integral x cos(3 x) dx

Answer 17

= 1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/4 integral x^2 sin(x) dx-1/6 x sin(3 x)+1/6 integral sin(3 x) dx

Answer 18

= 1/18 integral sin(u) du+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/4 integral x^2 sin(x) dx-1/6 x sin(3 x)

Answer 19

1/18 integral sin(u) du-3/4 x^2 cos(x)+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx-1/6 x sin(3 x)+3/2 integral x cos(x) dx

Answer 20

1/18 integral sin(u) du-3/4 x^2 cos(x)+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/2 x sin(x)-1/6 x sin(3 x)-3/2 integral sin(x) dx

Answer 21

= -(cos(u))/18-1/4 (3 x^2 cos(x))+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/2 x sin(x)-1/6 x sin(3 x)-3/2 integral sin(x) dx
= -(cos(u))/18-3/4 x^2 cos(x)+1/4 x^2 cos(3 x)+3/2 integral x^2 sin(x) dx+3/2 x sin(x)-1/6 x sin(3 x)+(3 cos(x))/2
= -(cos(u))/18-9/4 x^2 cos(x)+1/4 x^2 cos(3 x)+3/2 x sin(x)-1/6 x sin(3 x)+(3 cos(x))/2+3 integral x cos(x) dx
= -(cos(u))/18-9/4 x^2 cos(x)+1/4 x^2 cos(3 x)+9/2 x sin(x)-1/6 x sin(3 x)+(3 cos(x))/2-3 integral sin(x) dx
The integral of sin(x) is -cos(x):
= -(cos(u))/18-9/4 x^2 cos(x)+1/4 x^2 cos(3 x)+9/2 x sin(x)-1/6 x sin(3 x)+(9 cos(x))/2+constant
Substitute back for u = 3 x:
= -9/4 x^2 cos(x)+1/4 x^2 cos(3 x)+9/2 x sin(x)-1/6 x sin(3 x)+(9 cos(x))/2-1/18 cos(3 x)+constant
Which is equal to:
Answer: |
| = 1/36 (-81 (x^2-2) cos(x)+(9 x^2-2) cos(3 x)-6 x (sin(3 x)-27 sin(x)))+constant

Answer 22

I happen to did this before so im pretty sure its right .. but good luck,, check it w ur teacher , but if it wrong then please dont kill me ^^

Answer 23

oh my gosh! what a long integral!
thank you so much for help me out!, and be sure i wont kill u if its wrong :P :D ^^

Answer 24

^^ np

Answer 25

hmm, I did not read what you wrote but certainly no need for all that.

after letting x^3 = u,

=> 3x^2 dx= du

so integral becomes (sin u du)

=> -cos u + C
=>-cos(x^3) + C

as simple as that.