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

The integral of 1/(sqrt(4x-x^2)) is arcsin(1/2 x-1)+C

Popular >

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

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

arcsin (\frac{1}{2} x - 1) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{2} (x - 2))

Simplify

1 \cdot arcsin (\frac{1}{2} (x - 2)) : arcsin (\frac{1}{2} x - 1)

= arcsin (\frac{1}{2} x - 1)

Add a constant to the solution

= arcsin (\frac{1}{2} x - 1) + C

x^{2} - 4 \cdot x, x \cdot 0, [- 2, 4]

\frac{4}{x ^{6}} + \frac{2}{x} - \frac{5}{x ^{4}}

g (x) = 6 x^{2} e^{- x}

\frac{2 ( s + 2 )}{s ^{2} + s + 3}

x \to 0 + lim (\frac{1}{e ^{x}})

What is the integral of 1/(sqrt(4x-x^2)) ?
The integral of 1/(sqrt(4x-x^2)) is arcsin(1/2 x-1)+C