How do you evaluate the integral $\int \frac{1}{\sqrt{4 - x^{2}} d x}$ $int 1/(sqrt(4-x^2)dx$ from 0 to 2?

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How do you evaluate the integral $\int \frac{1}{\sqrt{4 - x^{2}} d x}$ $int 1/(sqrt(4-x^2)dx$ from 0 to 2?

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Answer 1 · 2016-08-24T12:25:07

$\frac{π}{2}$ $pi/2$ .

Explanation:

Knowing that, $\int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = a r c sin (\frac{x}{a}) + C$ $int1/sqrt(a^2-x^2)dx=arc sin(x/a)+C$ , we have,

$\int_{0}^{2} \frac{1}{\sqrt{4 - x^{2}}} d x = {[a r c sin (\frac{x}{2})]}_{0}^{2} = a r c sin 1 - a r c sin 0 = \frac{π}{2}$ $int_0^2 1/sqrt(4-x^2)dx=[arc sin(x/2)]_0^2=arc sin1-arc sin0=pi/2$ .

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How do you evaluate the integral $\int \frac{1}{\sqrt{4 - x^{2}} d x}$ $int 1/(sqrt(4-x^2)dx$ from 0 to 2?

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Explanation:

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How do you evaluate the integral $\int \frac{1}{\sqrt{4 - x^{2}} d x}$ $int 1/(sqrt(4-x^2)dx$ from 0 to 2?