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

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

Popular >

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

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

π

Decimal

3.14159 \dots

Solution steps

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

Complete the square

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

= \int_{0}^{4} \frac{1}{- ( x - 2 ) ^{2} + 4} d x

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v

Integral of a constant:

\int a d x = a x

= [1 \cdot v]_{- \frac{π}{2}}^{\frac{π}{2}}

Simplify

= [v]_{- \frac{π}{2}}^{\frac{π}{2}}

Compute the boundaries:

π

= π

\int_{0}^{1} x^{4} cos (x^{5}) d x

\int_{0}^{2} 4 x cos (x^{2}) d x

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

\int_{1}^{e^{5}} \frac{10 ln ( x )}{x} d x

\int_{0}^{3} (8 x - 1)^{2} d x

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