# How do I evaluate the indefinite integral intsin^2(2t)dt ?

Jul 28, 2014

$= \frac{1}{2} t - \frac{1}{8} \sin 4 t + c$, where $c$ is a constant

Explanation,

$= \int {\sin}^{2} \left(2 t\right) \mathrm{dt}$

Using trigonometric identity, $\cos 2 t = 1 - 2 {\sin}^{2} t$

${\sin}^{2} t = \frac{1 - \cos 2 t}{2}$, inserting this value of ${\sin}^{2} t$ in integral, we get

$= \int \frac{1 - \cos 4 t}{2} \mathrm{dt}$

$= \frac{1}{2} \int 1 \mathrm{dt} - \frac{1}{2} \int \cos 4 t \mathrm{dt}$

$= \frac{1}{2} t - \frac{1}{2} \frac{\sin 4 t}{4} + c$, where $c$ is a constant

$= \frac{1}{2} t - \frac{1}{8} \sin 4 t + c$, where $c$ is a constant