What is the integral of sqrt(-4x^2-12x-5) ?

The integral of sqrt(-4x^2-12x-5) is arcsin(x+3/2)+1/2 sin(2arcsin(x+3/2))+C

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integral of sqrt(-4x^2-12x-5)

\int - 4 x^{2} - 12 x - 5 d x

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

Solution steps

\int - 4 x^{2} - 12 x - 5 d x

Complete the square

- 4 x^{2} - 12 x - 5 : - 4 (x + \frac{3}{2})^{2} + 4

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

Apply u-substitution

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

Rewrite using trig identities

= 2 \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

= 2 \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

= 2 \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)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of sqrt(-4x^2-12x-5) ?
The integral of sqrt(-4x^2-12x-5) is arcsin(x+3/2)+1/2 sin(2arcsin(x+3/2))+C