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

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

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

\int_{1}^{3} \frac{1}{4 x - x ^{2}} d x

\frac{π}{3}

Decimal

1.04719 \dots

Solution steps

\int_{1}^{3} \frac{1}{4 x - x ^{2}} d x

Complete the square

4 x - x^{2} : - (x - 2)^{2} + 4

= \int_{1}^{3} \frac{1}{- ( x - 2 ) ^{2} + 4} d x

Apply u-substitution

: \int_{- 1}^{1} \frac{1}{- u ^{2} + 4} d u

= \int_{- 1}^{1} \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{6}}^{\frac{π}{6}} 1 d v

= \int_{- \frac{π}{6}}^{\frac{π}{6}} 1 d v

Integral of a constant:

\int a d x = a x

= [1 \cdot v]_{- \frac{π}{6}}^{\frac{π}{6}}

Simplify

= [v]_{- \frac{π}{6}}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{3}

= \frac{π}{3}

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