Question

If $I=\int \frac{dx}{2\sin x+\sec x}=\frac{1}{\sqrt{A}}\log |\text{cosec}(x+\frac{\pi }{4})-\cot (x+\frac{\pi }{4})|+\frac{1}{B(\sin x+\cos x)}+c$,then $A+B$ is

Answer 1

$I=\int \frac{dx}{2\sin x+\sec x}=\frac{\cos xdx}{2\sin x\cos x+1}\Rightarrow I=\frac{1}{2}\int \frac{2\cos xdx}{(\sin x+\cos x{)}^2}\Rightarrow I=\frac{1}{2}\int \frac{\cos x+\sin x}{(\sin x+\cos x{)}^2}dx+\frac{1}{2}\int \frac{\cos x-\sin x}{(\sin x+\cos x{)}^2}dx$
$\Rightarrow I=\frac{1}{2}\int \frac{1}{(\sin x+\cos x)}dx+\frac{1}{2}\int \frac{\cos x-\sin x}{(\sin x+\cos x{)}^2}dx$
Put $(\sin x+\cos x)=t\Rightarrow (\cos x-\sin x)dx=dt$
$\Rightarrow I=\frac{1}{2}\left(\int \frac{1}{\sqrt{2}}\times \frac{1}{\sin (x+\frac{\pi }{4})}dx+\int \frac{dt}{t^2}\right)$
$\Rightarrow I=\frac{1}{2}\left(\int \frac{1}{\sqrt{2}}\times \text{cosec}(x+\frac{\pi }{4})dx-\frac{1}{\sin x+\cos x}\right)$
$I=\frac{1}{2\sqrt{2}}\log |\text{cosec}(x+\frac{\pi }{4})-\cot (x+\frac{\pi }{4})|-\frac{1}{2(\sin x+\cos x)}+c$
$\Rightarrow A=8,B=-2\Rightarrow A+B=6$