What is the integral of sqrt(9-A^2) ?

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

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integral of sqrt(9-A^2)

\int 9 - A^{2} d A

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

Solution steps

\int 9 - A^{2} d A

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

= 9 \cdot \int \frac{1 + cos ( 2 u )}{2} d u

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 u) d u

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 u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

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

Substitute back

u = arcsin (\frac{1}{3} A)

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

Simplify

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

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

Add a constant to the solution

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

x = y^{2} - 2 y, x - y - 4 = 0

h^{^{'}}

\int cos^{3} (θ) d θ

4 y^{^{''}} + 4 y^{^{'}} + y = 0

\frac{d}{d x} (\frac{x ^{6}}{x ^{3}})

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