Question

How do you find the antiderivative of `sinxcosx`?

Answer 1

`-1/4 cos 2x + C`
or `1/2 sin^2 x + C`
or `-1/2 cos^2 x + C`

Explanation:

well, `sin x cos x = (sin 2x) /2` so you are looking at `1/2 int \ sin 2x \ dx = (1/2) [( 1/2) ( -cos 2x) + C] = -1/4 cos 2x + C'`

or maybe easier you can notice the pattern that `(sin^n x)' = n sin ^{n-1} x cos x` and pattern match. here `n-1 = 1` so n = 2 so we trial `(sin^2 x)'` which gives us `color{red}{2} sin x cos x` so we now that the anti deriv is `1/2 sin^2 x + C`

the other pattern also works ie `(cos^n x)' = n cos ^{n-1} x (-sin x) = - n cos ^{n-1} x sin x`

so trial solution `(-cos^2 x)' = -2 cos x (-sin x) = 2 cos x sin x` so the anti deriv is `-1/2 cos^2 x + C`