Primitive of Power of Tangent of a x by Square of Secant of a x

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Theorem

$\displaystyle \int \tan^n a x \sec^2 a x \rd x = \frac {\tan^{n + 1} a x} {\paren {n + 1} a} + C$


Proof

\(\displaystyle z\) \(=\) \(\displaystyle \tan a x\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d z} {\d x}\) \(=\) \(\displaystyle a \sec^2 a x\) Derivative of Tangent Function: Corollary
\(\displaystyle \implies \ \ \) \(\displaystyle \int \tan^n a x \sec^2 a x \rd x\) \(=\) \(\displaystyle \int \frac 1 a z^n \rd z\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 a \frac {z^{n + 1} } {n + 1}\) Primitive of Power
\(\displaystyle \) \(=\) \(\displaystyle \frac {\tan^{n + 1} a x} {\paren {n + 1} a} + C\) substituting for $z$ and simplifying

$\blacksquare$


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