What is the integral of sqrt(8-4x-4x^2) ?

The integral of sqrt(8-4x-4x^2) is 9/8 (2arcsin(2/3 x+1/3)+sin(2arcsin(2/3 x+1/3)))+C

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integral of sqrt(8-4x-4x^2)

\int 8 - 4 x - 4 x^{2} d x

\frac{9}{8} (2 arcsin (\frac{2}{3} x + \frac{1}{3}) + sin (2 arcsin (\frac{2}{3} x + \frac{1}{3}))) + C

Solution steps

\int 8 - 4 x - 4 x^{2} d x

Complete the square

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

= \int - 4 (x + \frac{1}{2})^{2} + 9 d x

Apply u-substitution

: \int - 4 u^{2} + 9 d u

= \int - 4 u^{2} + 9 d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= \frac{9}{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

= \frac{9}{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

= \frac{9}{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)

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

Substitute back

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

Simplify

\frac{9}{2} \cdot \frac{1}{2} (arcsin (\frac{2}{3} (x + \frac{1}{2})) + \frac{1}{2} sin (2 arcsin (\frac{2}{3} (x + \frac{1}{2})))) : \frac{9}{8} (2 arcsin (\frac{2}{3} x + \frac{1}{3}) + sin (2 arcsin (\frac{2}{3} x + \frac{1}{3})))

= \frac{9}{8} (2 arcsin (\frac{2}{3} x + \frac{1}{3}) + sin (2 arcsin (\frac{2}{3} x + \frac{1}{3})))

Add a constant to the solution

= \frac{9}{8} (2 arcsin (\frac{2}{3} x + \frac{1}{3}) + sin (2 arcsin (\frac{2}{3} x + \frac{1}{3}))) + C

x \to 0 lim (x^{2} + 4 x + 2)

n = 1 \sum \infty \frac{1}{ln ( n + 1 )}

y^{^{'}} = k (y - 9 - 10 cos (2 \cdot π \cdot t))

\frac{d ^{2}}{d x ^{2}} (2 sin (- 4 x - 3))

n = 2 \sum \infty \frac{1}{5 ^{n}}

What is the integral of sqrt(8-4x-4x^2) ?
The integral of sqrt(8-4x-4x^2) is 9/8 (2arcsin(2/3 x+1/3)+sin(2arcsin(2/3 x+1/3)))+C