Question

How do you integrate `int (2+sin(x/2))^2 cos(x/2)dx` using integration by parts?

Answer 1

`I = 10sin(x/2) -2cos(x) -2sin(x/2) + 2/3sin^3(x/2) + c`

Explanation:

For a non-parts approach

`I = int(2+sin(x/2))^2cos(x/2)dx`

Expanding that

`I = int(4+4sin(x/2)+sin^2(x/2))cos(x/2)dx`

And that

`I = int4cos(x/2)dx + int4sin(x/2)cos(x/2)dx + intsin^2(x/2)cos(x/2)dx`

The first integral is easy

`I = 8sin(x/2) + int4sin(x/2)cos(x/2)dx + intsin^2(x/2)cos(x/2)dx`

The second is easy if you remember that

`sin(alphax)cos(alphax) = sin(2alphax)/2`, so

`I = 8sin(x/2) + int2sin(x)dx + intsin^2(x/2)cos(x/2)dx`
`I = 8sin(x/2) -2cos(x) + intsin^2(x/2)cos(x/2)dx`

If you remember `sin^2(theta) = 1 -cos^2(theta)`

`I = 8sin(x/2) -2cos(x) + int(1 - cos^2(x/2))cos(x/2)dx`
`I = 8sin(x/2) -2cos(x) + intcos(x/2)dx - intcos^3(x/2)dx`
`I = 8sin(x/2) -2cos(x) + 2sin(x/2)- intcos^3(x/2)dx`

And now, use `cos^2(x/2) = 1 - sin^2(x/2)`

`I = 10sin(x/2) -2cos(x) - int(1-sin^2(x/2))cos(x/2)dx`

Say `u = sin(x/2)` so `du = cos(x/2)/2`

`I = 10sin(x/2) -2cos(x) -2 int(1-sin^2(x/2))cos(x/2)/2dx`
`I = 10sin(x/2) -2cos(x) -2 int(1-u^2)du`
`I = 10sin(x/2) -2cos(x) -2u + (2u^3)/3 + c`

Switching it back to `x`

`I = 10sin(x/2) -2cos(x) -2sin(x/2) + 2/3sin^3(x/2) + c`