Evaluate each of the following integral: `int_0^4 1/(sqrt(16-x^2))dx`

Explanation

Let


`I=\int_{0}^{4} \frac{1}{\sqrt{16-x^{2}}} d x`


Substituting `x=4 \sin \theta \Rightarrow d x=4 \cos \theta d \theta`



Also, When `x=0, \theta=0 and x=4, \theta=\frac{\pi}{2}`
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