Question

\[\int\limits_0^{\pi/2} \sqrt{1 + \sin x}\ dx\]

Answer 1

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