Question

$$\displaystyle\int{ \frac{ \sin (x-a) }{ \sin (x+b) } }d x$$

Answer

-(cos(x-a)*d*x)/sin(x+b)

Solution


Use Integration by Substitution on \(\int \sin{(x-a)} \, dx\).
Let \(u=x-a\), \(du=dx\)
Using \(u\) and \(du\) above, rewrite \(\int \sin{(x-a)} \, dx\).
\[\int \sin{u} \, du\]
Use Trigonometric Integration: the integral of \(\sin{u}\) is \(-\cos{u}\).
\[-\cos{u}\]
Substitute \(u=x-a\) back into the original integral.
\[-\cos{(x-a)}\]
Rewrite the integral with the completed substitution.
\[-\frac{(\cos{(x-a)})dx}{\sin{(x+b)}}\]