What is the integral of sqrt(12-x^2) ?

The integral of sqrt(12-x^2) is 6(arcsin(1/(2sqrt(3))x)+1/2 sin(2arcsin(1/(2sqrt(3))x)))+C

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integral of sqrt(12-x^2)

\int 12 - x^{2} d x

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

Solution steps

\int 12 - x^{2} d x

Apply Trigonometric Substitution

: \int 12 cos^{2} (u) d u

= \int 12 cos^{2} (u) d u

Take the constant out:

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

= 12 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 12 \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

= 12 \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

= 12 \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)

= 12 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

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

= 12 \cdot \frac{1}{2} (arcsin (\frac{1}{2 3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 3} x)))

Simplify

12 \cdot \frac{1}{2} (arcsin (\frac{1}{2 3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 3} x))) : 6 (arcsin (\frac{1}{2 3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 3} x)))

= 6 (arcsin (\frac{1}{2 3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 3} x)))

Add a constant to the solution

= 6 (arcsin (\frac{1}{2 3} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{2 3} x))) + C

7 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0

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

\int \frac{x}{( x - 1 )} d x

\frac{d}{d x} (\frac{1}{( ln ( x ) ) ^{2}})

\frac{d}{d x} (3 x^{8} ln (x))

What is the integral of sqrt(12-x^2) ?
The integral of sqrt(12-x^2) is 6(arcsin(1/(2sqrt(3))x)+1/2 sin(2arcsin(1/(2sqrt(3))x)))+C