# How do you integrate int sin^2x by integration by parts method?

Jul 26, 2016

$I = \frac{1}{2} \left(x - \sin x \cos x\right) + C$

#### Explanation:

so you have to use IBP on

$I = \int {\sin}^{2} x \setminus \mathrm{dx}$

$= \int \sin x \cdot \sin x \setminus \mathrm{dx}$

$= \int \sin x \cdot \frac{d}{\mathrm{dx}} \left(- \cos x\right) \setminus \mathrm{dx}$

which by IBP becomes...
$I = - \sin x \cos x + \int \frac{d}{\mathrm{dx}} \left(\sin x\right) \cdot \cos x \setminus \mathrm{dx}$

$= - \sin x \cos x + \int {\cos}^{2} x \setminus \mathrm{dx}$

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

$= - \sin x \cos x + x - I + C$

$I = \frac{1}{2} \left(x - \sin x \cos x\right) + C$