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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Evaluate the following integrals:
$\int_{0}^\limits{{2\pi}}\sqrt{1-\sin\frac{\text{x}}{2}}\text{ dx}$
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Answer

$\text{I}=\int_{0}^\limits{{2\pi}}\sqrt{1-\sin\frac{\text{x}}{2}}\text{ dx}$
$=\int_{0}^\limits{{2\pi}}\sqrt{\cos^2\frac{\text{x}}{4}+\sin^2\frac{\text{x}}{4}+2\sin\frac{\text{x}}{4}\cos\frac{\text{x}}{4}}\text{ dx}$
$=\int_{0}^\limits{{2\pi}}\sqrt{\Big(\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big)^2}\text{ dx}$
$=\int_{0}^\limits{{2\pi}}\Big[\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big]\text{dx}$
When $0\leq\text{x}\leq2\pi,0\leq\frac{\text{x}}{4}\leq\frac{\pi}{2}$
$\therefore\ \sin\frac{\text{x}}{4}\geq0,\cos\frac{\text{x}}{4}\geq0$
$\Rightarrow\Big[\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big]=\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}$
$\therefore\ \text{I}=\int_{0}^\limits{{2\pi}}\Big(\cos\frac{\text{x}}{4}+\sin\frac{\text{x}}{4}\Big)\text{dx}$
$=\Bigg[\frac{\sin\frac{\text{x}}{4}}{\frac{1}{4}}\Bigg]^{2\pi}_0+\Bigg[\frac{\big(-\cos\frac{\text{x}}{4}\big)}{\frac{1}{4}}\Bigg]^{2\pi}_0$
$=4\Big(\sin\frac{\pi}{2}-\sin0\Big)-4\Big(\cos\frac{\pi}{2}-\cos0\Big)$
$=4(1-0)-4(0-1)$
$=4+4$
$=8$

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