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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{x - x ^{2}} d x

Complete the square

x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{1}{4}

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \frac{1}{2} \cdot 2 \cdot \int \frac{1}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot 2 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{cos ( \frac{2 π}{x} )}{x ^{2}} d x

\int 2 y + 3 d y

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

\int \frac{1}{x ^{2} - 6 x + 11} d x

\int \frac{15 x ^{2} - 4 x - 28}{x ^{3} - 4 x} d x

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