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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

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

= \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

f (x) = - 8 x

\frac{d}{d x} (\frac{arccos ( x )}{( x + 1 ) ^{2}})

\frac{d}{d x} (e^{g (x)} sin (x))

\int_{- \infty}^{12} x e^{\frac{x}{4}} d x

\frac{d y}{d x}, 3 x^{2} + 2 y^{2} = 30

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