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Question
Solution
\[Let\ I = \int_0^\pi \frac{1}{1 + \sin x} d\ x\ . Then, \]
\[ I = \int_0^\pi \frac{1 - \sin x}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)} d x\]
\[ \Rightarrow I = \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x} dx \]
\[ \Rightarrow I = \int_0^\pi \frac{1 - \sin x}{\cos^2 x} dx \left[ \because \sin^2 x + \cos^2 x = 1 \right]\]
\[ \Rightarrow I = \int_0^\pi \sec^2 x - \sec x \tan x dx\]
\[ \Rightarrow I = \left[ \tan x - \sec x \right]_0^\pi \]
\[ \Rightarrow I = \left( \tan \pi - \sec \pi \right) - \left( \tan 0 - \sec 0 \right)\]
\[ \Rightarrow I = 0 + 1 - \left( 0 - 1 \right)\]
\[ \Rightarrow I = 1 + 1\]
\[ \Rightarrow I = 2\]
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