What is the integral of 3/(sqrt(4x-x^2)) ?

The integral of 3/(sqrt(4x-x^2)) is 3arcsin(1/2 x-1)+C

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integral of 3/(sqrt(4x-x^2))

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

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

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{4 x - x ^{2}} d x

Complete the square

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

= 3 \cdot \int \frac{1}{- ( x - 2 ) ^{2} + 4} d x

Apply u-substitution

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

= 3 \cdot \int \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int 1 d v

= 3 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 3 \cdot 1 \cdot v

Substitute back

= 3 \cdot 1 \cdot arcsin (\frac{1}{2} (x - 2))

Simplify

3 \cdot 1 \cdot arcsin (\frac{1}{2} (x - 2)) : 3 arcsin (\frac{1}{2} x - 1)

= 3 arcsin (\frac{1}{2} x - 1)

Add a constant to the solution

= 3 arcsin (\frac{1}{2} x - 1) + C

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What is the integral of 3/(sqrt(4x-x^2)) ?
The integral of 3/(sqrt(4x-x^2)) is 3arcsin(1/2 x-1)+C