# Primitive of x by Square of Tangent of a x

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## Theorem

$\displaystyle \int x \tan^2 a x \ \mathrm d x = \frac {x \tan a x} a + \frac 1 {a^2} \ln \left\vert{\cos a x}\right\vert - \frac {x^2} 2 + C$

## 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 \ \$ $\displaystyle \frac {\mathrm d u} {\mathrm d x}$ $=$ $\displaystyle 1$ Primitive of Power

and let:

 $\displaystyle \frac {\mathrm d v} {\mathrm d x}$ $=$ $\displaystyle \tan^2 a x$ $\displaystyle \implies \ \$ $\displaystyle v$ $=$ $\displaystyle \frac {\tan a x} a - x$ Primitive of $\tan^2 a x$

Then:

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

$\blacksquare$