Question

Evaluate the integral? : `int3xsinxcosx dx`?

Answer 1

`int \ 3xsinxcosx \ dx = 3/8 sin2x -3/4xcos2x + C`

Explanation:

We want to evaluate:

`I = int \ 3xsinxcosx \ dx`

First note that using the double angle formula for the sine function:

`sin 2A -= 2 sinAcosA`

We can write the integral as:

`I = 3/2 int \ (x)2sinxcosx \ dx`
`\ \ = 3/2 int \ xsin2x \ dx`

We can use Integration By Parts (IBP). Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

So for the integrand `xsin2x`, hopefully you can see that `x` simplifies when differentiated.

Let `{ (u,=x, => , (du)/dx=1), ((dv)/dx,=sin2x, =>, v=-1/2cos2x ) :}`

Then plugging into the IBP formula:

`int \ u(dv)/dx \ dx = uv - int \ v(du)/dx \ dx`

We get:

`int \ (x)(sin2x) \ dx = (x)(-1/2cos2x) - int \ (-1/2cos2x)(1) \ dx`

Hence:

`I = 3/2{-1/2xcos2x+int \ 1/2 cos2x \ dx }`
`\ \ = 3/2{-1/2xcos2x+1/4 sin2x } + C`
`\ \ = 3/8 sin2x -3/4xcos2x + C`