What is the integral of sqrt(7-4x-4x^2) ?

The integral of sqrt(7-4x-4x^2) is 2arcsin(1/(sqrt(2))(x+1/2))+sin(2arcsin(1/(sqrt(2))(x+1/2)))+C

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integral of sqrt(7-4x-4x^2)

\int 7 - 4 x - 4 x^{2} d x

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

Solution steps

\int 7 - 4 x - 4 x^{2} d x

Complete the square

7 - 4 x - 4 x^{2} : - 4 (x + \frac{1}{2})^{2} + 8

= \int - 4 (x + \frac{1}{2})^{2} + 8 d x

Apply u-substitution

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

Apply Trigonometric Substitution

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

Take the constant out:

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

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

Rewrite using trig identities

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

What is the integral of sqrt(7-4x-4x^2) ?
The integral of sqrt(7-4x-4x^2) is 2arcsin(1/(sqrt(2))(x+1/2))+sin(2arcsin(1/(sqrt(2))(x+1/2)))+C