Primitive of Power of Cosine of a x over Power of Sine of a x/Reduction of Both Powers

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Theorem

$\displaystyle \int \frac {\cos^m a x} {\sin^n a x} \rd x = \frac {-\cos^{m - 1} a x} {a \paren {n - 1} \sin^{n - 1} a x} - \frac {m - 1} {n - 1} \int \frac {\cos^{m - 2} a x} {\sin^{n - 2} a x} \rd x + C$


Proof

With a view to expressing the primitive in the form:

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

let:

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


and let:

\(\displaystyle \frac {\d v} {\d x}\) \(=\) \(\displaystyle \frac {\cos a x} {\sin^n a x}\)
\(\displaystyle \) \(=\) \(\displaystyle \sin^{-n} a x \cos a x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle v\) \(=\) \(\displaystyle \frac {\sin^{-n + 1} a x} {a \paren {-n + 1} }\) Primitive of $\sin^n a x \cos a x$
\(\displaystyle \) \(=\) \(\displaystyle \frac {-1} {a \paren {n - 1} \sin^{n - 1} a x}\)


Then:

\(\displaystyle \int \frac {\cos^m a x} {\sin^n a x} \rd x\) \(=\) \(\displaystyle \int \cos^{m - 1} a x \frac {\cos a x} {\sin^n a x} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\cos^{m - 1} a x} \paren {\frac {-1} {a \paren {n - 1} \sin^{n - 1} a x} }\) Integration by Parts
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \int \paren {\frac {-1} {a \paren {n - 1} \sin^{n - 1} a x} } \paren {-\paren {m - 1} a \cos^{m - 2} a x \sin a x } \rd x + C\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {-\cos^{m - 1} a x} {a \paren {n - 1} \sin^{n - 1} a x} - \frac {m - 1} {n - 1} \int \frac {\cos^{m - 2} a x} {\sin^{n - 2} a x} \rd x + C\) simplifying

$\blacksquare$


Also see


Sources