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integrating_sin2x

# Integrating sin² x

$$\int \sin^2{x}\,dx = \frac{x - \sin{x}\cos{x}}{2} + C$$

## Using the Pythagorean identity

Integrate by parts using $u = \sin{x}, du = \cos{x}$ and $v = -\cos{x}, dv = \sin{x}$:

\begin{align} & \int \sin{x}\sin{x}\,dx = -\sin{x}\cos{x} - \int -\cos{x}\cos{x}\,dx & \int \sin^2{x}\,dx = -\sin{x}\cos{x} + \int \cos^2{x}\,dx \end{align}

Add $\int \sin{x}^2\,dx$ to both sides:

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

Apply integral linearity:

$$2\int \sin^2{x}\,dx = -\sin{x}\cos{x} + \int \left(\cos^2{x} + \sin^2{x}\right)\,dx$$

Apply the Pythagorean identity and divide both sides by 2:

\begin{align} 2\int \sin^2{x}\,dx &= -\sin{x}\cos{x} + \int 1\,dx &= -\sin{x}\cos{x} + x + C \int \sin^2{x}\,dx &= \frac{x - \sin{x}\cos{x}}{2} + C \end{align}

## Using the double angle formula

<alert warning> Careful! This derivation differs from the one in Integrating cos² x by just a sign.</alert> Rewrite the integral using the double angle formula for the cosine:

$$\int \sin^2{x}\,dx = \int \frac{1 - \cos{2x}}{2}\,dx$$

Substitute $u = 2x, du = 2\,dx$:

\begin{align} \int \frac{1 - \cos{2x}}{2}\,dx &= \int \frac{1 - \cos{u}}{2}\frac{1}{2}\,du &= \frac{1}{4} \left(\int 1\,du - \int \cos{u}\,du\right) &= \frac{u - \sin{u} + C}{4} \end{align}

Substitute back and simplify:

\begin{align} \frac{u - \sin{u} + C}{4} &= \frac{2x - \sin{2x} + C}{4} &= \frac{x - \frac{\sin{2x}}{2}}{2} + C &= \frac{x - \sin{x}\cos{x}}{2} + C \end{align}