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

The integral from 1 to 2 of 1/(sqrt(4x-x^2)) is pi/6

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

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

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int_{- 1}^{0} \frac{1}{- u ^{2} + 4} d u

= \int_{- 1}^{0} \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int_{- \frac{π}{6}}^{0} 1 d v

= \int_{- \frac{π}{6}}^{0} 1 d v

Integral of a constant:

\int a d x = a x

= [1 \cdot v]_{- \frac{π}{6}}^{0}

Simplify

= [v]_{- \frac{π}{6}}^{0}

Compute the boundaries:

\frac{π}{6}

= \frac{π}{6}

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What is the integral from 1 to 2 of 1/(sqrt(4x-x^2)) ?
The integral from 1 to 2 of 1/(sqrt(4x-x^2)) is pi/6