The given function is \(\displaystyle \frac{2}{1 – \sin 2x} \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \frac{2}{1 – \sin 2x} \, dx \).
Solution:
First, let’s simplify the given function:
\( \begin{aligned} &\frac{2}{1 – \sin 2x} \\ &= \frac{2}{1 – \sin 2x} \cdot \frac{1 + \sin 2x}{1 + \sin 2x} \\ &= \frac{2(1 + \sin 2x)}{1 – \sin^2 2x} \\ &= \frac{2(1 + \sin 2x)}{\cos^2 2x} \\ &= 2 \sec^2 2x + 2 \sec 2x \tan 2x \\ \end{aligned} \)Now, the integral becomes:
\( \begin{aligned} &\int \left(2 \sec^2 2x + 2 \sec 2x \tan 2x\right) dx \\ &= 2 \cdot \frac{1}{2} \tan 2x + 2 \cdot \frac{1}{2} \sec 2x + C, \quad \text{C is a constant} \\ &= \tan 2x + \sec 2x + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{\int \frac{2}{1 – \sin 2x} \, dx = \tan 2x + \sec 2x + C} \)Please let me know in the comments if you find any errors in this solution.