What is the integral of sqrt(5-4x-x^2) ?

The integral of sqrt(5-4x-x^2) is 9/2 (arcsin(1/3 (x+2))+1/2 sin(2arcsin(1/3 (x+2))))+C

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integral of sqrt(5-4x-x^2)

integral

5 - 4 x - x^{2}

\frac{9}{2} (arcsin (\frac{1}{3} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x + 2)))) + C

Solution steps

\int 5 - 4 x - x^{2} d x

Complete the square

5 - 4 x - x^{2} : - (x + 2)^{2} + 9

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

Apply u-substitution

= \int - u^{2} + 9 d u

Apply Trigonometric Substitution

= \int 9 cos^{2} (v) d v

Take the constant out:

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

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

Rewrite using trig identities

= 9 \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) 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 \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= 9 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x + 2))))

Simplify

9 \cdot \frac{1}{2} (arcsin (\frac{1}{3} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x + 2)))) : \frac{9}{2} (arcsin (\frac{1}{3} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x + 2))))

= \frac{9}{2} (arcsin (\frac{1}{3} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x + 2))))

Add a constant to the solution

= \frac{9}{2} (arcsin (\frac{1}{3} (x + 2)) + \frac{1}{2} sin (2 arcsin (\frac{1}{3} (x + 2)))) + C

What is the integral of sqrt(5-4x-x^2) ?
The integral of sqrt(5-4x-x^2) is 9/2 (arcsin(1/3 (x+2))+1/2 sin(2arcsin(1/3 (x+2))))+C