Question

$\int \frac{\cos x+\sqrt{3}}{1+4\sin (x+\frac{\pi }{3})+4{\sin }^2(x+\frac{\pi }{3})}dx$ is
where $c$ is constant of integration

• A. $\frac{\cos x}{1+2\sin (x+\frac{\pi }{3})}+c$
• B. $\frac{\sec x}{1+2\sin (x+\frac{\pi }{3})}+c$
• C. $\frac{\sin x}{1+2\sin (x+\frac{\pi }{3})}+c$
• D. $\frac{1}{2}{\tan }^{-1}(1+2\sin (x+\frac{\pi }{3}))+c$

Answer 1

The correct option is C $\frac{\sin x}{1+2\sin (x+\frac{\pi }{3})}+c$

$I=\int \frac{\cos x+\sqrt{3}}{1+4\sin (x+\frac{\pi }{3})+4{\sin }^2(x+\frac{\pi }{3})}dx$
$=\int \frac{\cos x+\sqrt{3}}{{(1+2\sin (x+\frac{\pi }{3}))}^2}dx$
$=\int \frac{\cos x+\sqrt{3}}{{(1+\sin x+\sqrt{3}\cos x)}^2}dx$
$=\int \frac{\cos x+\sqrt{3}}{{\sin }^{2}x{(\text{cosec}x+1+\sqrt{3}\cot x)}^2}dx$
$=\int \frac{\cot x\text{cosec}x+\sqrt{3}{\text{cosec}}^{2}x}{{(1+\text{cosec}x+\sqrt{3}\cot x)}^2}dx$

Put $1+\text{cosec}x+\sqrt{3}\cot x=t$
$\Rightarrow -(\text{cosec}x\cot x+\sqrt{3}{\text{cosec}}^{2}x)dx=dt$
$\therefore I=\int -\frac{dt}{t^2}$
$=\frac{1}{t}+c$
$=\frac{1}{1+\text{cosec}x+\sqrt{3}\cot x}+c$
$=\frac{\sin x}{1+2\sin (x+\frac{\pi }{3})}+c$