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integrating_cos2x

# Integrating cos² x

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

## Using the Pythagorean identity

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

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

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

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

Apply integral linearity:

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

Apply the Pythagorean identity and divide both sides by 2:

\begin{align} 2\int \cos^2{x}\,dx &= \sin{x}\cos{x} + \int 1\,dx &= \sin{x}\cos{x} + x + C \int \cos^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 sin² x by just a sign.</alert>

Rewrite the integral using the double angle formula for the cosine:

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

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}

integrating_cos2x.txt · Last modified: 2018/05/01 09:35 (external edit)