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Question
Solution
\[\int_0^4 \frac{1}{\sqrt{16 - x^2}} d x\]
\[ = \int_0^4 \frac{1}{\sqrt{4^2 - x^2}} d x\]
\[ = \left[ \sin^{- 1} \frac{x}{4} \right]_0^4 \]
\[ = \left( \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{2}\]
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