Question

How do you integrate `int x^3 sin^2 x dx` using integration by parts?

Answer 1

See the answer below:

Answer 2

`int \ x^3sin^2x \ dx = (2 x^4+(6x-4x^3)sin2x (3-6x^2)cos2x)/16 + C`

Explanation:

We seek:

`I = int \ x^3sin^2x \ dx`

In preparation for an application of integration by Parts, we note that using the identity `cos2A-=1-2sin^2A`:

`int \ sin^2x \ dx = 1/2 int \ 1-cos2x \ dx`
`" " = 1/2 (x-1/2sin2x)`
`" " = 1/2 x-1/4sin2x`

We can then apply Integration By Parts:

Let `{ (u,=x^3, => (du)/dx,=3x^2), ((dv)/dx,=sin^2 x, => v,=1/2 x-1/4sin2x ) :}`

Then plugging into the IBP formula:

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

We have:

`int \ x^3sin^2x \ dx = x^3(1/2 x-1/4sin2x) - int \ (1/2 x-1/4sin2x)3x^2 \ dx`
`:. I = 1/2 x^4-1/4x^3sin2x - 3/4 int \ 2 x^3-x^2sin2x \ dx`
`\ \ \ \ \ \ \ = 1/2 x^4-1/4x^3sin2x - 3/4 x^4/2+3/4 int \ x^2sin2x \ dx`
`\ \ \ \ \ \ \ = 1/8 x^4-1/4x^3sin2x +3/4 int \ x^2sin2x \ dx` ..... [A]

Now consider the last integral:

`I_1 = int \ x^2sin2x \ dx`

We can again apply integration by parts:

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

So we have:

`I_1 = (x^2)(-1/2cos2x) - int \ (-1/2cos2x)(2x) \ dx`
`\ \ \ = -1/2x^2cos2x + int \ xcos2x \ dx`..... [B]

And, now we have to consider:

`I_2 = int \ xcos2x \ dx`

Which again requires an application of Integration By Parts:

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

And so:

`I_2 = (x)(1/2sin2x) - int \ 1/2sin2x \ dx`
`\ \ \ = 1/2xsin2x + 1/4cos2x`..... [C]

Combining the result [B] and [C] with [A] we have:

`I = 1/8 x^4-1/4x^3sin2x +3/4 {-1/2x^2cos2x + 1/2xsin2x + 1/4cos2x} +C`
`\ \ = 1/16(2 x^4-4x^3sin2x) +3/16 {-2x^2cos2x + 2xsin2x + cos2x} +C`
`\ \ = (2 x^4-4x^3sin2x -6x^2cos2x + 6xsin2x + 3cos2x)/16 +C`
`\ \ = (2 x^4+(6x-4x^3)sin2x (3-6x^2)cos2x)/16 +C`