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Definite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals2 Marks
Question
Evaluate the following integrals:$\int\limits^4_0\frac{1}{\sqrt{16-\text{x}^2}}\text{ dx}$
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Answer

$\int\limits^4_0\frac{1}{\sqrt{16-\text{x}^2}}\text{ dx}$$=\int\limits^4_0\frac{1}{\sqrt{14^2-\text{x}^2}}\text{ dx}$
$=\Big[\sin^{-1}\frac{\text{x}}{4}\Big]^4_0$
$=\Big(\frac{\pi}{2}-0\Big)$
$=\frac{\pi}{2}$

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