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Question
Solution
\[Let\ I = \int_0^\frac{\pi}{2} \sqrt{1 + \cos x}\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \sqrt{1 + \cos x} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\sqrt{1 - \cos^2 x}}{\sqrt{1 - \cos x}} dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{\sin x}{\sqrt{1 - \cos x}} dx\]
\[Let 1 - \cos x = u\]
\[ \Rightarrow \sin x\ dx\ = du\]
\[ \therefore I = \int\frac{du}{\sqrt{u}}\]
\[ \Rightarrow I = \left[ 2\sqrt{u} \right]\]
\[ \Rightarrow I = \left[ 2\sqrt{1 - \cos x} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = 2 - 0\]
\[ \Rightarrow I = 2\]
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