Question

How do you find the indefinite integral of `int sin(5x)sin(8x)dx`?

Answer 1

`1/6sin3x - 1/26sin13x+C`

Explanation:

`sinxsiny=1/2(cos(x-y)-cos(x+y))`

`I=int sin5x sin8x dx=1/2 int (cos(-3x) - cos(13x)) dx`

`I=1/2 int cos3xdx - 1/2 int cos13x dx`

`I= 1/6sin3x - 1/26sin13x+C`

Answer 2

Let's ignore the integral symbols for now.

`=> sin(5x)sin(8x)`

It'd be great if we had an identity for this. Maybe we do!

`sinusinv = ?`

Recall that an addition/subtraction identity with `cos(upmv)` contains `sinusinv`.

`cos(u - v) = cosucosv + sinusinv`
`cos(u + v) = cosucosv - sinusinv`

We can acquire `2sinusinv` from this by subtracting the two equations (and thus subtracting `-sinusinv` from `sinusinv`):

`cos(u - v) - cos(u + v)`
`= [cosucosv + sinusinv] - [cosucosv - sinusinv]`

`= 2sinusinv`

Therefore:

`color(green)(sinusinv = (cos(u-v) - cos(u + v))/2)`

Applying this identity, we get:

`= [cos(5x - 8x) - cos(5x + 8x)]/2`

`= [cos(-3x) - cos(13x)]/2`

but `cos(3x) = cos(-3x)`, thus:

`= [cos(3x) - cos(13x)]/2`

This is much easier to do! Bringing back the integral symbols:

`= 1/2int cos(3x)dx - 1/2int cos(13x)dx`

`= 1/2*1/3sin(3x) - 1/2*1/13sin(13x)`

`= color(blue)(1/6sin(3x) - 1/26sin(13x) + C)`