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

Mar 27, 2015

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

$\int {\left[f \left(x\right)\right]}^{n} \cdot f ' \left(x\right) \mathrm{dx} = {\left[f \left(x\right)\right]}^{n + 1} / \left(n + 1\right) + c$,

so:

$\int \sin x \cos x \mathrm{dx} = {\sin}^{2} \frac{x}{2} + c$.

Mar 27, 2015

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

Let $u = 1$ and $\mathrm{dv} = \sin x \cos x \mathrm{dx}$

Then $\mathrm{du} = 0 \mathrm{dx}$ and $v = \int \sin x \cos x \mathrm{dx}$

This integral can be found (in two ways) using substitution. Let $w = \cos x$ so $\mathrm{dw} = - \sin x \mathrm{dx}$

With this substitution we get:

$\int \sin x \cos x \mathrm{dx} = - \frac{1}{2} {\cos}^{2} x + C$

So our integral by parts becomes:
$\int \left(1\right) \left(\sin x \cos x \mathrm{dx}\right) = \left(1\right) \left(- \frac{1}{2} {\cos}^{2} x\right) - \int \left(- \frac{1}{2} {\cos}^{2} x\right) \left(0\right) \mathrm{dx}$

$= - \frac{1}{2} {\cos}^{2} x + C$