What is the integral from 0 to 2 of 1/(sqrt(4-x^2)) ?

The integral from 0 to 2 of 1/(sqrt(4-x^2)) is pi/2

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integral from 0 to 2 of 1/(sqrt(4-x^2))

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x

\frac{π}{2}

Solution steps

\int_{0}^{2} \frac{1}{4 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 1 d u

= \int_{0}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{2}}

Simplify

= [u]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

(e^{y} + 6 x) d x + x e^{y} d y = 0

x \to \frac{1}{2} lim (4 x^{2} - 8 x + 5)

\int \frac{2}{1 - ( 2 x ) ^{2}} d x

y = 5 x^{8} e^{x}

\int (2 x + 3 x^{\frac{1}{2}}) x^{\frac{1}{3}} d x

What is the integral from 0 to 2 of 1/(sqrt(4-x^2)) ?
The integral from 0 to 2 of 1/(sqrt(4-x^2)) is pi/2