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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Take the constant out:

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 1 d v

= 2 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{( 1 )}{x ^{2}} d x

\int (7 x^{3}) d x

\int \frac{( 4 - x )}{( x ^{2} - 2 x - 3 ) ^{3}} d x

\int cos (x) - sin (x) + 3 d x

\int cos (π) x cos (4 π) x d x

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