Question
Evaluate the following definite integrals:
$\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$

Answer

Let $\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\text{ dx}$ Then,
$\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\sqrt{1+\sin\text{x}}\times\frac{\sqrt{1-\sin\text{x}}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{1-\sin^2\text{x}}}\text{ dx}$
$\Rightarrow\text{I}=\int_{0}^\limits{\frac{\pi}{2}}\frac{\cos\text{x}}{\sqrt{1-\sin\text{x}}}\text{ dx}$
Let $1-\sin\text{x}=\text{u}$
$\Rightarrow-\cos\text{x dx}=\text{du}$
$\therefore\ \text{I}=\int\frac{-\text{du}}{\sqrt{\text{u}}}$
$\Rightarrow\text{I}=\big[-2\sqrt{\text{u}}\big]$
$\Rightarrow\text{I}=\big[-2\sqrt{1-\sin\text{x}}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=0+2$
$\Rightarrow\text{I}=2$

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