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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the definite integral in Exercise:
$\int^{\frac{\pi}{3}}\limits_{\frac{\pi}{6}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin2\text{x}}}\text{dx}$
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Answer

$\text{Let I}=\int^{\frac{\pi}{3}}\limits_{\frac{\pi}{6}}\frac{\sin\text{x}+\cos\text{x}}{\sqrt{\sin2\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int^{\frac{\pi}{3}}\limits_{\frac{\pi}{6}}\frac{(\sin\text{x}+\cos\text{x)}}{\sqrt{- ( - \sin 2 \text{x})}}\text{dx}$
$\Rightarrow\text{I}=\int^{\frac{\pi}{3}}\limits_{\frac{\pi}{6}}\frac{(\sin\text{x}+\cos\text{x)}}{\sqrt{-(-1+1-2\sin\text{x}\cos\text{x})}}\text{dx}$
$\Rightarrow\text{I}=\int^{\frac{\pi}{3}}\limits_{\frac{\pi}{6}}\frac{(\sin\text{x}+\cos\text{x)}}{\sqrt{1-(\sin^{2}\text{x}\cos^{2}\text{x}-2\sin\text{x}\cos\text{x})}}\text{dx}$
$\Rightarrow\text{I}=\int^{\frac{\pi}{3}}\limits_{\frac{\pi}{6}}\frac{(\sin\text{x}+\cos\text{x)}\text{dx}}{\sqrt{1-(\sin\text{x}-\cos\text{x)}^{2}}}$
$\text{Let}(\sin\text{x}-\cos\text{x})=\text{t}\Rightarrow(\sin\text{x}+\cos\text{x)}\text{dx}=\text{dt}$
When $\text{x}=\frac{\pi}{6},\text{t}=\bigg(\frac{1-\sqrt{3}}{2}\bigg)$ and when $\text{x}=\frac{\pi}{3},\text{t}=\bigg(\frac{\sqrt{3}-1}{2}\bigg)$
$\text{I}=\int^{\frac{\sqrt{3}-1}{2}}_{\frac{1-\sqrt{3}}{2}}\frac{\text{dt}}{\sqrt{1-\text{t}^{2}}}$
$\Rightarrow\text{I}=\int^{\frac{\sqrt{3}-1}{2}}_{-\bigg(\frac{1-\sqrt{3}}{2}\bigg)}\frac{\text{dt}}{\sqrt{1-\text{t}^{2}}}$
As $\frac{1}{\sqrt{1-(-\text{t})}^{2}}=\frac{1}{\sqrt{1-\text{t}}^{2}}$, therefore, $\frac{1}{\sqrt{1-\text{t}^{2}}}$ is an even function.
It knot is wn that if $\text{f (x)}$is an even function on, then $\int^{\text{a}}_{\text{-a}}\text{f (x)}\ \text{dx}=2\int^{\text{a}}_{0}\text{f (x)}\text{dx}$
$\Rightarrow\text{I}=2\int^{\frac{\sqrt{3}-1}{2}}_{0}\frac{\text{dt}}{\sqrt{1-\text{t}^{2}}}$
$=\Big[2\sin^{-1}\text{t}\Big]^{\frac{\sqrt{3}-1}{2}}_{0}$

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