Question

$Evaluate\int \frac{\sin x}{\sin 3x}dx.$

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

Answer 1

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

$I=\int \frac{\sin x}{\sin 3x}=\int \frac{\sin x}{3sinx-4{\sin }^{3}x}dx$
$I=\int \frac{1}{3-4{\sin }^{2}x}dx$
$I=\int \frac{{\sec }^{2}x}{3{\sec }^{2}x-4{\tan }^{2}x}dx\left[Dividing{N}^{\prime }and{D}^{\prime }by{\cos }^{2}x\right]$
$Substitute\tan x=tand{\sec }^{2}xdx=dt,weget$
$I=\int \frac{dt}{3(1+t^2)-4t^2}=\int \frac{dt}{3-t^2}=\int \frac{1}{{\left(\sqrt{3}\right)}^2-t^2}dt$
$I=\int \frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C=\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x}\right|+C$