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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals2 Marks
Question
Find the value of $\int_0^{\frac{\pi}{2}} \sqrt{1+\sin x} d x$.
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Answer

Let
$
\begin{aligned}
I & =\int_0^{\frac{\pi}{2}} \sqrt{1+\sin x} d x \\
& =\int_0^{\frac{\pi}{2}} \sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}} d x \\
& =\int_0^{\frac{\pi}{2}}\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\
& =\left[-\frac{\cos \frac{x}{2}}{1 / 2}+\frac{\sin \frac{x}{2}}{1 / 2}\right]_0^{\frac{\pi}{2}} \\
& =2\left[\sin \frac{x}{2}-\cos \frac{x}{2}\right]_0^{\frac{\pi}{2}}
\end{aligned}
$
$\begin{array}{l}=2\left[\left(\sin \frac{\pi}{4}-\cos \frac{\pi}{4}\right)-(\sin 0-\cos 0)\right] \\ =2\left[\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)-(0-1)\right] \\ =2 \text { Ans. }\end{array}$

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