Question

Quick question.

Answer 1

\[\int\limits_{}{}x^2\sin(4x)dx\]

Answer 2

Alright, so for this problem, you're going to want to use the substitution method since you have multiple x-variables, if that makes sense.
U-substitution, it's called
so we set 4x = u

Answer 3

\[\int_{}{}x^2\sin(u)\]
u=4x
Right?

Answer 4

Right, and then you need to be aware of the derivative of u, which is 4

Answer 5

\[\int_{}{}x^2dx \times \int_{}{}\sin(u)....d?\]

Answer 6

Perhaps I am getting ahead of myself.

Answer 7

Is that correct?

Answer 8

Integrate by parts twice.

Answer 9

Forgive me, but I am crap at integration.

Answer 10

Or use tabular method (which is the same thing). Use:

u = x^2; dv = Sin(4x)dx

Answer 11

Ok, perhaps you could run me through the steps. I know....
\[\int_{}{}udv = uv - \int_{}{}vdu\]

Answer 12

ok so find v and du

Answer 13

\[\int\limits uv dx=1st*\int\limits 2nd dx-\int\limits differential of 1st*\int\limits 2nd dx *dx\]

Answer 14

du = 2x
v = (-cos(4x))/4
\[\int_{}{}x^2\sin(4x) = (x^2)(\frac{-\cos(4x)}{4}) - \int_{}{}(\frac{-\cos(4x)}{4})(2x)\]

Is this correct?

Answer 15

correct write dx at the end

Answer 16

So now I need to evaluate that integral.... joy...

Answer 17

1/2xcos4x dx

Answer 18

\[= (x^2)(\frac{-\cos(4x)}{4}) - \frac{-cos(4x) + 4x\sin(4x)}{32}\]
Is this correct?

Answer 19

\[\frac{ 1 }{ 2 }\int\limits x \cos 4x dx=\frac{ 1 }{ 2 }[x*\frac{ -\sin 4x }{4 }-\int\limits 1*\frac{ -\sin 4x }{ 4}dx+c\]
again integrate sin4x