Primitive of Reciprocal of Power of Cosine of a x

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Theorem

$\ds \int \frac {\d x} {\cos^n a x} = \frac {\sin a x} {a \paren {n - 1} \cos^{n - 1} a x} + \frac {n - 2} {n - 1} \int \frac {\d x} {\cos^{n - 2} a x}$


Proof

With a view to expressing the primitive in the form:

$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$

let:

\(\ds u\) \(=\) \(\ds \frac 1 {\cos^{n - 2} a x}\)
\(\ds \) \(=\) \(\ds \cos^{- n + 2} a x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds -a \paren {-n + 2} \cos^{-n + 1} a x \sin a x\) Derivative of $\cos a x$, Derivative of Power, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {a \paren {n - 2} \sin a x} {\cos^{n - 1} a x}\) simplifying


and let:

\(\ds \frac {\d v} {\d x}\) \(=\) \(\ds \frac 1 {\cos^2 a x}\)
\(\ds \) \(=\) \(\ds \sec^2 a x\) Secant is $\dfrac 1 \cos$
\(\ds \leadsto \ \ \) \(\ds v\) \(=\) \(\ds \frac {\tan a x} a\) Primitive of $\sec^2 a x$


Then:

\(\ds \int \frac {\d x} {\cos^n a x}\) \(=\) \(\ds \int \frac {\d x} {\cos^{n - 2} a x \cos^2 a x}\)
\(\ds \) \(=\) \(\ds \paren {\frac 1 {\cos^{n - 2} a x} } \paren {\frac {\tan a x} a} - \int \paren {\frac {\tan a x} a} \paren {\frac {a \paren {n - 2} \sin a x} {\cos^{n - 1} a x} } \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds \frac {\sin a x} {a \cos^{n - 1} a x} - \int \frac {\paren {n - 2} \sin^2 a x} {\cos^n a x} \rd x\) Tangent is $\dfrac {\sin} {\cos}$
\(\ds \) \(=\) \(\ds \frac {\sin a x} {a \cos^{n - 1} a x} - \int \frac {\paren {n - 2} \paren {1 - \cos^2 a x} } {\cos^n a x} \rd x\) Sum of $\sin^2$ and $\cos^2$
\(\ds \) \(=\) \(\ds \frac {\sin a x} {a \cos^{n - 1} a x} - \paren {n - 2} \int \frac {\d x} {\cos^n a x} + \paren {n - 2} \int \frac {\d x} {\cos^{n - 2} a x}\) Linear Combination of Primitives
\(\ds \leadsto \ \ \) \(\ds \paren {n - 1} \int \frac {\d x} {\cos^n a x}\) \(=\) \(\ds \frac {\sin a x} {a \cos^{n - 1} a x} + \paren {n - 2} \int \frac {\d x} {\cos^{n - 2} a x}\) gathering terms
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {\cos^n a x}\) \(=\) \(\ds \frac {\sin a x} {a \paren {n - 1} \cos^{n - 1} a x} + \frac {n - 2} {n - 1} \int \frac {\d x} {\cos^{n - 2} a x}\) dividing by $n - 1$

$\blacksquare$


Also see


Sources

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