Question

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

Answer 1

\[Let\ I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{1 + \sin x} d x . Then, \]
\[I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} d x\]
\[ \Rightarrow I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1 - \sin x}{1 - \sin^2 x} dx\]
\[ \Rightarrow I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1 - \sin x}{\cos^2 x} dx\]
\[ \Rightarrow I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx\]
\[ \Rightarrow I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \left( \sec^2 x - \sec x \tan x \right) dx\]
\[ \Rightarrow I = \left[ \tan x - \sec x \right]_{- \frac{\pi}{4}}^\frac{\pi}{4} \]
\[ \Rightarrow I = \left( 1 - \sqrt{2} \right) - \left( - 1 - \sqrt{2} \right)\]
\[ \Rightarrow I = 2\]