# How do you find the integral of sin^2(3x)dx?

Jul 12, 2016

$= \frac{1}{2} x - \frac{1}{12} \sin 6 x + C$

#### Explanation:

$\int \setminus {\sin}^{2} \left(3 x\right) \setminus \mathrm{dx}$

small book keeping gesture is to make the sub $u = 3 x , \mathrm{du} = 3 \mathrm{dx}$

$\frac{1}{3} \int \setminus {\sin}^{2} \left(u\right) \setminus \mathrm{du}$

then we use the cosine double angle formulae

$\cos 2 A = 1 - 2 {\sin}^{2} A$

so ${\sin}^{2} A = \frac{1 - \cos 2 A}{2}$

$= \frac{1}{6} \int \setminus 1 - \cos 2 u \setminus \mathrm{du}$

$= \frac{1}{6} \left(u - \frac{1}{2} \sin 2 u\right) + C$

$= \frac{1}{6} \left(3 x - \frac{1}{2} \sin \left(2 \cdot 3 x\right)\right) + C$

$= \frac{1}{2} x - \frac{1}{12} \sin 6 x + C$