(i) \(\int \frac{dx}{sin(x \,+\,a)sin(x\,+\,b)}\) = \(\frac{1}{sin(a\,-\,b)}\int\frac{sin(a\,-\,b)dx}{sin(x\,+\,a)sin(x\,+\,b)}\)
= \(\frac{1}{sin(a\,-\,b)}\int\frac{sin[(x\,+\,a)-(x\,+\,b)]dx}{sin(x\,+\,a)sin(x\,+\,b)}\)
= \(\frac{1}{sin(a\,-\,b)}\int\frac{sin(x\,+\,a)cos(x\,+\,b)\,-\,cos(x\,+\,a)sin(x\,+\,b)}{sin(x\,+\,a)sin(x\,+\,b)}dx\)
= \(\frac{1}{sin(a\,-\,b)}\int[cot(x+b)-cot(x+a)]dx\)
= \(\frac{1}{sin(a\,-\,b)}(log|sin(x+b)|-log|sin(x+a)|+c\)
(\(\because\) \(\int cot x \,dx=log|sinx|dx \) )
= \(\frac{1}{sin(a\,-\,b)} \,log\left|\frac{sin(x+b)}{sin(x+a)}\right|+c\) ( \(\because\) log A - log B = log \(\frac AB\))
(ii) \(\int \frac{dx}{sin(x+a)cos(x-b)}\) = \(\frac1{cos(a+b)}\int\frac{cos(a+b) dx}{sin(x+a)cos(x-b)}\)
= \(\frac1{cos(a+b)}\int\frac{cos((x+a)-(x-b))dx}{sin(x+a)cos(x-b)}\)
= \(\frac1{cos(a+b)}\int\frac{cos(x+a)cos(x-b) \,+\, sin(x+a)sin(x-b)}{sin(x+a)cos(x-b)}dx\)
= \(\frac1{cos(a+b)}\int[cot(x+a)\,+\,tan(x-b)]dx\)
= \(\frac1{cos(a+b)}[log|sin(x+a)|\,+\,log|sec(x-b)|]+c\)
= \(\frac1{cos(a+b)}\,log\left|\frac{sin(x+a)}{sec(x-b)}\right|+c\)
= \(\frac1{cos(a+b)}\,log|sin(x+a)cos(x-b)+c\)
(iii) \(\int \frac{dx}{cos(x-a)cos(x+b)}\) = \(\frac1{sin(a+b)}\int\frac{sin(a+b)}{cos(x-a)cos(x+b)}dx\)
= \(\frac1{sin(a+b)}\int\frac{sin((x+b)\,-\,(x-a))\,dx}{cos(x-a)cos(x+b)}\)
= \(\frac1{sin(a+b)}\int\frac{sin(x+b)cos(x-a)\,-\,cos(x+b)sin(x-a)}{cos(x-a)cos(x+b)}\)
= \(\frac1{sin(a+b)}\int [tan(x+b)-tan(x-a)]dx\)
= \(\frac1{sin(a+b)}[log|sec(x+b)|-log|sec(x-a)|] +c\)
= \(\frac1{sin(a+b)}\,log \left|\frac{sec(x+b)}{sec(x-a)}\right|+c\)
= \(\frac1{sin(a+b)}\,log \left|\frac{cos(x-a)}{cos(x+b)}\right|+c\).