Question

\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals

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Answer 1

2

 

\[\int_0^\pi \frac{1}{1 + \sin x} d x\]
\[ = \int_0^\pi \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x}dx\]
\[ = \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x}dx\]
\[ = \int_0^\pi \frac{1 - \sin x}{\cos^2 x}dx\]
\[ = \int_0^\pi \left( se c^2 x - \sec x \tan x \right) dx\]
\[ = \left[ \tan x - sec x \right]_0^\pi \]
\[ = 0 + 1 - 0 + 1\]
\[ = 2\]