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

The integral from 1 to 2 of 1/(sqrt(4-x^2)) is pi/3

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

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

\frac{π}{3}

Decimal Notation

1.04719 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

= \int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{\frac{π}{6}}^{\frac{π}{2}}

Simplify

= [u]_{\frac{π}{6}}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{3}

= \frac{π}{3}

\frac{d}{d x} ((5 x + 7) x^{2})

\frac{d}{d x} (e^{2 x} cos (x))

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

5 e^{8 t + 9}

\frac{2 x ^{3} + 5}{x}

What is the integral from 1 to 2 of 1/(sqrt(4-x^2)) ?
The integral from 1 to 2 of 1/(sqrt(4-x^2)) is pi/3