Question

$\int \frac{dx}{4{\sin }^{2}x+3{\cos }^{2}x}$ is equal to :

• A. $\frac{\sqrt{3}}{4}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{3}}\right)+C$
• B. $\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\tan x}{\sqrt{3}}\right)+C$
• C. $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{3}}\right)+C$
• D. $\frac{\sqrt{3}}{2}{\tan }^{-1}\left(\frac{\tan x}{\sqrt{3}}\right)+C$
• E. $\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{3}}\right)+C$

Answer 1

The correct option is E $\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{3}}\right)+C$

Let $I=\int \frac{dx}{4{\sin }^{2}x+3{\cos }^{3}x}$

Dividing numerator and denominator by ${\cos }^{2}x,$ we get

$=\int \frac{{\sec }^{2}xdx}{4{\tan }^{2}x+3}$

Put $\tan x=t$

$\Rightarrow {\sec }^{2}xdx=dt$

$\therefore$, $I=\int \frac{dt}{4t^2+3}=\frac{1}{4}\int \frac{1}{\left(t^2+\frac{3}{4}\right)}dt$

$=\frac{1}{4}\times \frac{1}{\sqrt{3}/2}{\tan }^{-1}\left(\frac{t}{\sqrt{3}/2}\right)+C$

$=\frac{1}{2\sqrt{3}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{3}}\right)+C$