# How do you find the Integral from 0 to #(3sqrt(3))/2# for #sqrt(9-x^2)#?

Try this (olha para a substituição que é bastante notável!):

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To find the integral of ( \sqrt{9 - x^2} ) from ( 0 ) to ( \frac{3\sqrt{3}}{2} ), you can use trigonometric substitution.

Let ( x = 3\sin(\theta) ), then ( dx = 3\cos(\theta) d\theta ).

When ( x = 0 ), ( \theta = 0 ). When ( x = \frac{3\sqrt{3}}{2} ), ( \theta = \frac{\pi}{3} ).

Now, the integral becomes:

[ \int_{0}^{\frac{\pi}{3}} 3\cos(\theta) \cdot 3\cos(\theta) d\theta ]

[ = 9\int_{0}^{\frac{\pi}{3}} \cos^2(\theta) d\theta ]

[ = \frac{9}{2} \left(\theta + \frac{\sin(2\theta)}{2}\right) \Bigg|_{0}^{\frac{\pi}{3}} ]

[ = \frac{9}{2} \left(\frac{\pi}{3} + \frac{\sqrt{3}}{2}\right) ]

[ = \frac{9\pi}{6} + \frac{9\sqrt{3}}{4} ]

[ = \frac{3\pi}{2} + \frac{3\sqrt{3}}{2} ]

So, the integral from ( 0 ) to ( \frac{3\sqrt{3}}{2} ) for ( \sqrt{9 - x^2} ) is ( \frac{3\pi}{2} + \frac{3\sqrt{3}}{2} ).

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