What is the integral of sqrt(-x^2-6x-5) ?

The integral of sqrt(-x^2-6x-5) is 2arcsin(1/2 (x+3))+sin(2arcsin(1/2 (x+3)))+C

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integral of sqrt(-x^2-6x-5)

\int - x^{2} - 6 x - 5 d x

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

Solution steps

\int - x^{2} - 6 x - 5 d x

Complete the square

- x^{2} - 6 x - 5 : - (x + 3)^{2} + 4

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int 4 cos^{2} (v) d v

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

Take the constant out:

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

= 4 \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

= 4 \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

= 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} (x + 3)) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} (x + 3))))

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sqrt(-x^2-6x-5) ?
The integral of sqrt(-x^2-6x-5) is 2arcsin(1/2 (x+3))+sin(2arcsin(1/2 (x+3)))+C