Theorem: Primitive of Power of Cosine of a x over Power of Sine of a x/Reduction of Power of Cosine

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

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:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac {\cos^{m - 1} a x} {\sin^n a x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = \frac {\sin^n a x \dfrac \d {\d x} \cos^{m - 1} a x - \cos^{m - 1} a x \dfrac \d {\d x} \sin^n a x} {\sin^{2 n} a x}
      | c = Quotient Rule for Derivatives
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \frac \d {\d x} \cos^{m - 1} a x
      | r = a \paren {m - 1} \cos^{m - 2} a x \paren {-\sin a x}
      | c = Derivative of $\cos a x$, Derivative of Power, Chain Rule for Derivatives
}}
{{eqn | r = -a \paren {m - 1} \cos^{m - 2} a x \sin a x
      | c = 
}}
{{eqn | l = \frac \d {\d x} \sin^n a x
      | r = a n \sin^{n - 1} a x \cos a x
      | c = Derivative of $\sin a x$, Derivative of Power, Chain Rule for Derivatives
}}
{{end-eqn}}
and so:
{{begin-eqn}}
{{eqn | l = \frac {\d u} {\d x}
      | r = \frac {\sin^n a x \paren {-a \paren {m - 1} \cos^{m - 2} a x \sin a x} - \cos^{m - 1} a x \paren {a n \sin^{n - 1} a x \cos a x} } {\sin^{2 n} a x}
      | c = 
}}
{{eqn | r = \frac {-a \paren {m - 1} \cos^{m - 2} a x \sin^2 a x + a n \cos^m a x} {\sin^{n + 1} a x}
      | c = simplifying and cancelling $\sin^{n - 1}$
}}
{{eqn | r = \frac {-a \cos^{m - 2} a x \paren {\paren {m - 1} \sin^2 a x + n \cos^2 a x} } {\sin^{n + 1} a x}
      | c = factorising
}}
{{eqn | r = \frac {-a \cos^{m - 2} a x \paren {\paren {m - 1} \paren {1 - \cos^2 a x} + n \cos^2 a x} } {\sin^{n + 1} a x}
      | c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \frac {-a \cos^{m - 2} a x \paren {\paren {n - m + 1} \cos^2 a x + \paren {m - 1} } } {\sin^{n + 1} a x}
      | c = simplifying
}}
{{eqn | r = \frac {-a \paren {n - m + 1} \cos^m a x} {\sin^{n + 1} a x} + \frac {a \paren {m - 1} \cos^{m - 2} a x} {\sin^{n + 1} a x}
      | c = separating
}}
{{end-eqn}}
Then let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \cos a x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {\sin a x} a
      | c = Primitive of $\cos a x$
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\cos^m a x} {\sin^n a x} \rd x
      | r = \int \frac {\cos^{m - 1} a x} {\sin^n a x} \cos^a x \rd x 
      | c = 
}}
{{eqn | r = \paren {\frac {\cos^{m - 1} a x} {\sin^n a x} } \paren {\frac {\sin a x} a}
      | c = Integration by Parts
}}
{{eqn | o = 
      | ro= -
      | r = \int \paren {\frac {\sin a x} a} \paren {\frac {-a \paren {n - m + 1} \cos^m a x} {\sin^{n + 1} a x} + \frac {a \paren {m - 1} \cos^{m - 2} a x} {\sin^{n + 1} a x} } \rd x + C
      | c = 
}}
{{eqn | r = \frac {\cos^{m - 1} a x} {a \sin^{n - 1} a x}
      | c = 
}}
{{eqn | o = 
      | ro= +
      | r = \paren {n - m + 1} \int \frac {\cos^m a x} {\sin^n a x} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | o = 
      | ro= +
      | r = \paren {m - 1} \int \frac {\cos^{m - 2} a x} {\sin^n a x} \rd x + C
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | ll= \leadsto
      | l = \paren {1 - \paren {n - m + 1} } \int \frac {\cos^m a x} {\sin^n a x} \rd x
      | r = \frac {\cos^{m - 1} a x} {a \sin^{n - 1} a x} + \paren {m - 1} \int \frac {\cos^{m - 2} a x} {\sin^n a x} \rd x + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {m - n} \int \frac {\cos^m a x} {\sin^n a x} \rd x
      | r = \frac {\cos^{m - 1} a x} {a \sin^{n - 1} a x} + \paren {m - 1} \int \frac {\cos^{m - 2} a x} {\sin^n a x} \rd x + C
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\cos^m a x} {\sin^n a x} \rd x
      | r = \frac {\cos^{m - 1} a x} {a \paren {m - n} \sin^{n - 1} a x} + \frac {m - 1} {m - n} \int \frac {\cos^{m - 2} a x} {\sin^n a x} \rd x + C
      | c = 
}}
{{end-eqn}}
{{qed}}