Why don't you try something new?

$\Rightarrow \text{I} = \int \dfrac{1}{1 + \sin 2x} \text{dx}$ $\Rightarrow \text{I} = \int \dfrac{1}{1 + \cos\left( \dfrac{\pi}{2} - 2x\right)} \text{dx}$ $\Rightarrow \text{I} = \int \dfrac{1}{2\cos^2\left(\dfrac{\pi}{4} - x\right)}\text{dx}$ $\Rightarrow \text{I} = \dfrac{1}{2} \int \sec^2\left( \dfrac{\pi}{4} - x\right) \text{dx}$ $\Rightarrow \text{I} = \dfrac{1}{2}\tan\left( x - \dfrac{\pi}{4}\right) + \ C$
$\Large \color{#3D99F6}{OR}$
$\Rightarrow \text{I} = \int \dfrac{1}{1 + \sin 2x}\text{dx}$ $\Rightarrow \text{I} = \int \dfrac{1}{1 + \dfrac{2\tan x}{1 + \tan^2x}}\text{dx}$ $\Rightarrow \text{I} = \int \dfrac{1 + \tan^2 x}{(1 + \tan x)^2}\text{dx}$ $\text{substituting}\ \tan x = t \Rightarrow \sec^2x \text{dx} = \text{dt}$ $\Rightarrow \text{I} = \dfrac{-1}{1 + \tan x} + \ D$ $\text{We can add}\ \dfrac{1}{2} \ \text{to our integration as it's a contant}$ $\Rightarrow \text{I} = \dfrac{-1}{1 + \tan x} + \dfrac{1}{2} + \ C$ $\Rightarrow \text{I} = \dfrac{1}{2}\tan\left( x - \dfrac{\pi}{4}\right) + \ C$

Writing 1 + $\sin 2x$ as { $\sin x$ + $\cos x$ }^2 and then simplifying it to a single cosine term simplifies the integral greatly.