Question

How do you find the integral of `sinxcos xdx` by using integration by parts?

Answer 1

It's not a good idea di solve it by parts, because it is an immediate integral, using this rule:

`int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c`,

so:

`intsinxcosxdx=sin^2x/2+c`.

Answer 2

I would prefer to integrate this by substitution rather than parts. But if you insist:

Let `u=1` and `dv=sin x cos x dx`

Then `du=0 dx` and `v= int sin x cos x dx`

This integral can be found (in two ways) using substitution. Let `w=cosx` so `dw=- sin x dx`

With this substitution we get:

`int sin x cos x dx = -1/2 cos^2 x +C`

So our integral by parts becomes:
`int (1) (sin x cos x dx) = (1)(-1/2 cos^2x) - int (-1/2 cos^2x) (0) dx`

`= -1/2 cos^2x +C`