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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
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Answer

Let $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$ $\Bigg[\int\limits^{\text{b}}_{\text{a}}\text{f(x)}\text{dx}=\int\limits^{\text{b}}_{\text{a}}\text{f}(\text{a}+\text{b}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\sin\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\cos\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}}+\sqrt{\sin\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}{\sqrt{\sin\text{x}}+\sqrt{\cos\text{x}}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}$
$\Rightarrow2\text{I}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_{\frac{\pi}{6}}$
$\Rightarrow2\text{I}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
$\Rightarrow\text{I}=\frac{\pi}{12}$

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