# What is int 16sin^2 xcos^2 x dx ?

Jan 7, 2016

$2 x - \sin \frac{4 x}{2} + k$ with $k \in \mathbb{R}$.

#### Explanation:

We have to remember a few formulas. Here, we will need $2 \sin \left(\theta\right) \cos \left(\theta\right) = \sin \left(2 \theta\right)$. We can make it appear easily because we're dealing with the squares of $\sin \left(x\right)$ and $\cos \left(x\right)$ and we're multiplying them by an even number.

$16 {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right) = 4 \left(4 {\cos}^{2} \left(x\right) {\sin}^{2} \left(x\right)\right) = 4 {\left(2 \sin \left(x\right) \cos \left(x\right)\right)}^{2} = 4 {\left(\sin \left(2 x\right)\right)}^{2}$.

So $\int 16 {\sin}^{2} \left(x\right) {\cos}^{2} \left(x\right) \mathrm{dx} = 4 \int {\sin}^{2} \left(2 x\right) \mathrm{dx}$.

And we know that ${\sin}^{2} \left(\theta\right) = \frac{1 - \cos \left(2 \theta\right)}{2}$ because $\cos \left(2 \theta\right) = 1 - 2 {\sin}^{2} \left(\theta\right)$, so ${\sin}^{2} \left(2 x\right) = \frac{1 - \cos \left(4 x\right)}{2}$.

Hence the final result : $4 \int {\sin}^{2} \left(2 x\right) = 4 \int \frac{1 - \cos \left(4 x\right)}{2} \mathrm{dx} = 4 \int \frac{\mathrm{dx}}{2} - 4 \int \cos \frac{4 x}{2} \mathrm{dx} = 2 x - 2 \int \cos \left(4 x\right) \mathrm{dx} = 2 x + c - 2 \sin \frac{4 x}{4} + a$ with $a , c \in \mathbb{R}$. Let's say $k = a + c$, hence the final answer.