Question

If $I=\int \frac{\sin (x+\alpha )+\cos x}{\sin (x-\alpha )}dx$, then I equals

• A. $2\cos \alpha (1+\sin \alpha )\log |\sin (x-\alpha )|+(\cos 2\alpha +\cos \alpha )x+C$
• B. $\cos \alpha (1+2\sin \alpha )\log |\sin (x-\alpha )|+(\cos 2\alpha -\sin \alpha )x+C$
• C. $2\sin \alpha (1+\cos \alpha )\log |\sin (x-\alpha )|+(\cos 2\alpha +\cos \alpha )x+C$
• D. None of these

Answer 1

The correct option is B $\cos \alpha (1+2\sin \alpha )\log |\sin (x-\alpha )|+(\cos 2\alpha -\sin \alpha )x+C$

Let $I=\int \frac{\sin (x+\alpha )+\cos x}{\sin (x-\alpha )}dx$

Substitute $(x-\alpha )=\theta$

$I=\int \frac{\sin (\theta +2\alpha )+\cos (\theta +\alpha )}{\sin \theta }d\theta$

$=\int \frac{\sin \theta \cos 2\alpha +\cos \theta \sin 2\alpha +\cos \theta \cos \alpha -\sin \theta \sin \alpha }{\sin \theta }d\theta$

$=\int \left\{\left(\sin 2\alpha +\cos \alpha \right)\cot \theta +\left(\cos 2\alpha -\sin \alpha \right)\right\}d\theta$
$=\left(\sin 2\alpha +\cos \alpha \right)\log \left|\sin \theta \right|+\left(\cos 2\alpha -\sin \alpha \right)\theta +c$

$=(2\sin \alpha \cos \alpha +\cos \alpha )\log |\sin (x-\alpha )|+(\cos 2\alpha -\sin \alpha )x-(\cos 2\alpha -\sin \alpha )\alpha +c$

$=\cos \alpha (1+2\sin \alpha )\log |\sin (x-\alpha )|+(\cos 2\alpha -\sin \alpha )x+C$