What is the integral of sqrt(6-x^2) ?

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

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integral of sqrt(6-x^2)

\int 6 - x^{2} d x

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

Solution steps

\int 6 - x^{2} d x

Apply Trigonometric Substitution

: \int 6 cos^{2} (u) d u

= \int 6 cos^{2} (u) d u

Take the constant out:

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

= 6 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 6 \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

= 6 \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

= 6 \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)

= 6 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{6} x)

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

Simplify

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

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

Add a constant to the solution

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

y^{^{'}} = \frac{1 - y ^{2}}{t}

f (x) = 4 x^{2} + 2 x, (- 2, 12)

\int \frac{x ^{7}}{2} - \frac{1}{x ^{5}} d x

n \to \infty lim (sin (2 n))

e^{- 3 x}, 0

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