Question

Evaluate integrate 1/sinx (2+cosx) dx.

Answer 1

The question is not clear if the term with 2+cos x is in the numerator or denominator.  I assume it is in numerator.

$I= \int\limits^{}_{} {\frac{2+Cos\ x}{Sin\ x}} \, dx = 2 \int\limits^{}_{} {Cosec\ x} \, dx + \int\limits^{}_{} {Cot\ x} \, dx\\\\ I1=\int\limits^{}_{} {Cot\ x} \, dx=Ln | Sin\ x | \\\\ I2= \int\limits^{}_{} {Cosec\ x} \, dx\\\\Let\ tan\ \frac{x}{2}=t,\ \ dt=\frac{1}{2}(1+t^2)dx,\ Cosec\ x=\frac{1}{Sinx}=\frac{1+t^2}{2t}\\\\ I2= \int\limits^{}_{} {\frac{1+t^2}{2t}*\frac{2}{1+t^2}} \, dt\\\\ I2= \int\limits^{}_{} {\frac{1}{t}} \, dt =Ln\ | tan\ \frac{x}{2} | \\\\ I=2Ln |tan\ \frac{x}{2}|+Ln|Sin\ x|+K$