2014/074) integrate $\dfrac{\sin\ x}{a \cos \ x + b\sin\ x} $

Back ground information
$∫ \dfrac{\sin\ x}{\sin \ x + \ cos\ x} dx $
= $∫\dfrac{1}{2} \dfrac{2\sin\ x}{\sin \ x + \ cos\ x} dx $
= $∫\dfrac{1}{2} \dfrac{(\sin\ x+\cos \ x) +(\sin\ x-\cos \ x) }{\sin \ x + \ cos\ x} dx $
= $∫\dfrac{1}{2} (1+ \dfrac{\sin\ x-\cos \ x }{\sin \ x + \ cos\ x} dx $
=  $∫\dfrac{1}{2}dx + \dfrac{1}{2} \dfrac{\sin\ x-\cos \ x }{\sin \ x + \ cos\ x} dx $
= $\dfrac{1}{2}x - \dfrac{1}{2} ln \mid (\sin \ x + \ cos\ x)\mid $

as $\dfrac{\cos\ x}{\sin \ x + \ cos\ x}  $
=  $1-  \dfrac{\sin\ x}{\sin \ x + \ cos\ x} $

so $∫ \dfrac{\cos \ x}{\sin \ x + \ cos\ x} dx $
=  $\dfrac{1}{2}x + \dfrac{1}{2} ln \mid (\sin \ x + \ cos\ x)\mid $
Now integral

now we need to inetgrate
$\dfrac{\sin\ x}{a \cos \ x + b\ sin\  x} $

as derivate of  $a \cos \ x + b\sin \  x $ is $- a \sin  \ x + b\cos \   x $

we can put  $\sin\ x= m(a \cos \ x + b\ sin\  x) + n (- a \sin  \ x + b\cos \   x) $

comparing coefficients we get 1 = ma + nb and 0=mb-na

solving these 2 we get

$m=\dfrac{b}{b^2+a^2}$ and $n=  \dfrac{- a}{b^2+a^2}$

so  $\sin\ x=\dfrac{b}{b^2+a^2}(a \cos \ x + b\ sin\  x) - \dfrac{a}{b^2+a^2} (- a \sin  \ x + b\cos \   x) $

so $\dfrac{\sin\ x}{a \ cos \ x + b\sin \ x }= \dfrac{b}{b^2+a^2} - \dfrac{a}{b^2+a^2} \dfrac{- a \sin  \ x + b\cos \   x} {a \cos \ x + b\sin \ x } $

so integral of above = $ \dfrac{b}{b^2+a^2} x - \dfrac{a}{b^2+a^2} ln \mid a \cos \ x + b\sin \ x \mid $