# Primitive of Sine of a x over Sine of a x minus Cosine of a x

## Theorem

$\displaystyle \int \frac {\sin a x \rd x} {\sin a x - \cos a x} = \frac x 2 + \frac 1 {2 a} \ln \size {\sin a x - \cos a x} + C$

## Proof

First note that:

 $\text {(1)}: \quad$ $\displaystyle \map {\frac {\d} {\d x} } {\sin a x - \cos a x}$ $=$ $\displaystyle a \paren {\cos a x + \sin a x}$ Derivative of $\sin a x$ and Derivative of $\cos a x$

Then:

 $\displaystyle \int \frac {\sin a x \rd x} {\sin a x - \cos a x}$ $=$ $\displaystyle \int \frac {\paren {\sin a x - \cos a x + \cos a x} \rd x} {\sin a x - \cos a x}$ $\displaystyle$ $=$ $\displaystyle \int \frac {\paren {\sin a x - \cos a x} \rd x} {\sin a x - \cos a x} + \int \frac {\cos a x \rd x} {\sin a x - \cos a x}$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \int \rd x + \int \frac {\cos a x \rd x} {\sin a x - \cos a x}$ simplification $\displaystyle$ $=$ $\displaystyle \int \rd x + \int \frac {\paren {\cos a x + \sin a x - \sin a x} \rd x} {\sin a x - \cos a x}$ $\displaystyle$ $=$ $\displaystyle \int \rd x + \int \frac {\paren {\cos a x + \sin a x} \rd x} {\sin a x - \cos a x} - \int \frac {\sin a x \rd x} {\sin a x - \cos a x}$ Linear Combination of Integrals $\displaystyle \leadsto \ \$ $\displaystyle 2 \int \frac {\sin a x \rd x} {\sin a x - \cos a x}$ $=$ $\displaystyle \int \rd x + \int \frac {\paren {\cos a x + \sin a x} \rd x} {\sin a x - \cos a x}$ rearranging $\displaystyle \leadsto \ \$ $\displaystyle \int \frac {\sin a x \rd x} {\sin a x + \cos a x}$ $=$ $\displaystyle \frac 1 2 \int \rd x + \frac 1 {2 a} \int \frac {a \paren {\cos a x + \sin a x} \rd x} {\sin a x - \cos a x}$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \frac x 2 + \frac 1 {2 a} \int \frac {a \paren {\cos a x + \sin a x} \rd x} {\sin a x - \cos a x} + C$ Primitive of Constant Then from $(1)$: $\displaystyle$ $=$ $\displaystyle \frac x 2 + \frac 1 {2 a} \ln \size {\sin a x - \cos a x} + C$ Primitive of Function under its Derivative

$\blacksquare$