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

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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

\int_{0}^{1} (5 e^{x} - 5 x e^{x^{2}}) d x

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

\frac{d}{d x} (x^{- 2 x})

x \to 0 lim (x^{8} cos (\frac{5}{x}))

2 y y^{^{'}} + 4 = y^{2} + 4 x, y (0) = 9

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