Primitive of Power of Tangent of a x by Square of Secant of a x
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Theorem
- $\ds \int \tan^n a x \sec^2 a x \rd x = \frac {\tan^{n + 1} a x} {\paren {n + 1} a} + C$
Proof
\(\ds z\) | \(=\) | \(\ds \tan a x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d z} {\d x}\) | \(=\) | \(\ds a \sec^2 a x\) | Derivative of $\tan a x$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \tan^n a x \sec^2 a x \rd x\) | \(=\) | \(\ds \int \frac 1 a z^n \rd z\) | Integration by Substitution | ||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \frac {z^{n + 1} } {n + 1}\) | Primitive of Power | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\tan^{n + 1} a x} {\paren {n + 1} a} + C\) | substituting for $z$ and simplifying |
$\blacksquare$
Also see
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $\tan a x$: $14.432$