What is the integral of (9x^2)/(sqrt(4x-x^2)) ?

The integral of (9x^2)/(sqrt(4x-x^2)) is 9(6arcsin(1/2 x-1)-sin(2arcsin(1/2 x-1))-4sqrt(-x^2+4x))+C

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integral of (9x^2)/(sqrt(4x-x^2))

\int \frac{9 x ^{2}}{4 x - x ^{2}} d x

9 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x) + C

Solution steps

\int \frac{9 x ^{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

= 9 \cdot \int \frac{x ^{2}}{4 x - x ^{2}} d x

Complete the square

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

= 9 \cdot \int \frac{x ^{2}}{- ( x - 2 ) ^{2} + 4} d x

Apply u-substitution

= 9 \cdot \int \frac{( u + 2 ) ^{2}}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

= 9 \cdot \int (2 sin (v) + 2)^{2} d v

Expand

(2 sin (v) + 2)^{2} : 4 sin^{2} (v) + 8 sin (v) + 4

= 9 \cdot \int 4 sin^{2} (v) + 8 sin (v) + 4 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (\int 4 sin^{2} (v) d v + \int 8 sin (v) d v + \int 4 d v)

\int 4 sin^{2} (v) d v = 2 v - sin (2 v)

\int 8 sin (v) d v = - 8 cos (v)

\int 4 d v = 4 v

= 9 (2 v - sin (2 v) - 8 cos (v) + 4 v)

Substitute back

= 9 (2 arcsin (\frac{1}{2} (x - 2)) - sin (2 arcsin (\frac{1}{2} (x - 2))) - 8 cos (arcsin (\frac{1}{2} (x - 2))) + 4 arcsin (\frac{1}{2} (x - 2)))

Simplify

9 (2 arcsin (\frac{1}{2} (x - 2)) - sin (2 arcsin (\frac{1}{2} (x - 2))) - 8 cos (arcsin (\frac{1}{2} (x - 2))) + 4 arcsin (\frac{1}{2} (x - 2))) : 9 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x)

= 9 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x)

Add a constant to the solution

= 9 (6 arcsin (\frac{1}{2} x - 1) - sin (2 arcsin (\frac{1}{2} x - 1)) - 4 - x^{2} + 4 x) + C

What is the integral of (9x^2)/(sqrt(4x-x^2)) ?
The integral of (9x^2)/(sqrt(4x-x^2)) is 9(6arcsin(1/2 x-1)-sin(2arcsin(1/2 x-1))-4sqrt(-x^2+4x))+C