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integral of sin^2(x)

What is the integral of sin^2(x)?

The integral of \sin^2(x) is \frac{1}{2}(x - \sin(x)\cos(x)) + C.

Proof. We want to determine the integral of \sin^2(x), that is:
\begin{align*}
\int \sin^2(x) dx.
\end{align*}
We have seen here the equality \sin^2(x) = \frac{1}{2}(1 - \cos(2x)). Furthermore, we know from here that \sin(2x) = 2\sin(x)\cos(x). Therefore, we have that:
\begin{align*}
\int \sin^2(x) dx &= \int \frac{1}{2}(1 - \cos(2x)) dx \\
&= \frac{1}{2} \int (1 - \cos(2x)) dx, \quad \sin^2(x) = \frac{1}{2}(1 - \cos(2x)) \\
&= \frac{1}{2} \bigg(x - \frac{1}{2}\sin(2x)\bigg) + C \\
&= \frac{x}{2} - \frac{1}{4}\sin(2x) + C \\
&= \frac{x}{2} - \frac{1}{4}\cdot 2\sin(x)\cos(x) + C, \quad  \sin(2x) = 2\sin(x)\cos(x) \\
&= \frac{1}{2}(x - \sin(x)\cos(x)) + C.
\end{align*}
So, the integral of \sin^2(x) is \frac{1}{2}(x - \sin(x)\cos(x)) + C.

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