Primitive of x by Square of Cosine of a x

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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

$\blacksquare$


Also see


Sources