# Primitive of x by Square of Cosine of a x

## Theorem

$\displaystyle \int x \cos^2 a x \ \mathrm d x = \frac {x^2} 4 + \frac {x \sin 2 a x} {4 a} + \frac {\cos 2 a x} {8 a^2} + C$

## Proof

With a view to expressing the primitive in the form:

$\displaystyle \int u \frac {\mathrm d v}{\mathrm d x} \ \mathrm d x = u v - \int v \frac {\mathrm d u}{\mathrm d x} \ \mathrm d x$

let:

 $\displaystyle u$ $=$ $\displaystyle x$ $\displaystyle \implies \ \$ $\displaystyle \frac {\mathrm d u}{\mathrm d x}$ $=$ $\displaystyle 1$ Derivative of Identity Function

and let:

 $\displaystyle \frac {\mathrm d v}{\mathrm d x}$ $=$ $\displaystyle \cos^2 a x$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle \frac x 2 + \frac {\sin 2 a x} {4 a}$ Primitive of $\cos^2 a x$

Then:

 $\displaystyle \int x \cos^2 a x \ \mathrm d x$ $=$ $\displaystyle x \left({\frac x 2 + \frac {\sin 2 a x} {4 a} }\right) - \int \left({\frac x 2 + \frac {\sin 2 a x} {4 a} }\right) \times 1 \ \mathrm d x + C$ Integration by Parts $\displaystyle$ $=$ $\displaystyle \frac {x^2} 2 + \frac {x \sin 2 a x} {4 a} - \frac 1 2 \int x \ \mathrm d x - \frac 1 {4 a} \int \sin 2 a x \ \mathrm d x + C$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \frac {x^2} 2 + \frac {x \sin 2 a x} {4 a} - \frac {x^2} 4 - \frac 1 {4 a} \int \sin 2 a x \ \mathrm d x + C$ Primitive of Power $\displaystyle$ $=$ $\displaystyle \frac {x^2} 2 + \frac {x \sin 2 a x} {4 a} - \frac {x^2} 4 - \frac 1 {4 a} \frac {\left({-\cos 2 a x}\right)} {2 a} + C$ Primitive of $\sin a x$ $\displaystyle$ $=$ $\displaystyle \frac {x^2} 4 + \frac {x \sin 2 a x} {4 a} + \frac {\cos 2 a x} {8 a^2} + C$ simplifying

$\blacksquare$