Question

\[\int\frac{2}{2 + \sin 2x}\text{ dx }\]

Answer 1

\[\text{ Let I } = \int \frac{2}{2 + \sin \left( 2x \right)}\text{ dx }\]
\[ = \int \frac{2}{2 + 2 \sin x \cos x}\text{ dx }\]
\[ = \int \frac{1}{1 + \sin x \cos x}\text{ dx }\]
\[\text{Dividing numerator and denominator by} \cos^2 x\]
\[ \Rightarrow I = \int \frac{\sec^2 x \text{ dx }}{\sec^2 x + \tan x}\]
\[ = \int \frac{\sec^2 x \text{ dx}}{1 + \tan^2 x + \tan x}\]
\[\text{ Let tan x }= t\]
\[ \Rightarrow \sec^2 \text{ x }dx = dt\]
\[ \therefore I = \int \frac{dt}{t^2 + t + 1}\]
\[ = \int\frac{dt}{t^2 + t + \frac{1}{4} - \frac{1}{4} + 1}\]
\[ = \int \frac{dt}{\left( t + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}\]
\[ = \frac{2}{\sqrt{3}} \text{ tan }^{- 1} \left( \frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C\]
\[ = \frac{2}{\sqrt{3}} \text{ tan }^{- 1} \left( \frac{2t + 1}{\sqrt{3}} \right) + C\]
\[ = \frac{2}{\sqrt{3}} \text{ tan }^{- 1} \left( \frac{2 \tan x + 1}{\sqrt{3}} \right) + C\]