Question

The value of $\int \frac{\sin x}{\sin 3x}dx$ is
(where $C$ is integration constant)

• A. $\frac{1}{\sqrt{3}}\ln \left|\frac{\sqrt{3}-\tan x}{\sqrt{3}+\tan x}\right|+C$
• B. $\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C$
• C. $\frac{1}{\sqrt{3}}\ln \left|\frac{2\sqrt{3}+\tan x}{2\sqrt{3}-\tan x}\right|+C$
• D. $\frac{1}{2\sqrt{3}}\ln \left|\frac{2\sqrt{3}-\tan x}{2\sqrt{3}+\tan x}\right|+C$

Answer 1

The correct option is B $\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C$

$I=\int \frac{\sin x}{\sin 3x}dx=\int \frac{\sin x}{3\sin x-4{\sin }^{3}x}dx=\int \frac{1}{3-4{\sin }^{2}x}dx(\because \sin x\ne 0)$
Dividing numerator and denominator by ${\cos }^{2}x$,
$\Rightarrow I=\int \frac{{\sec }^{2}x}{3{\sec }^{2}x-4{\tan }^{2}x}dx$
Putting $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt$
Now,
$I=\int \frac{dt}{3(1+t^2)-4t^2}=\int \frac{dt}{3-t^2}=\int \frac{1}{{\left(\sqrt{3}\right)}^2-t^2}dt=\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C[\because \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\ln \left|\frac{a+x}{a-x}\right|+C]=\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C$