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

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

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integral of sqrt(4-2x^2)

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 22 cos^{2} (u) d u

= \int 22 cos^{2} (u) d u

Take the constant out:

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

= 22 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 22 \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

= 22 \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

= 22 \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)

= 22 \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

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

= 22 \frac{1}{2} (arcsin (\frac{2}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{2}{2} x)))

Simplify

22 \frac{1}{2} (arcsin (\frac{2}{2} x) + \frac{1}{2} sin (2 arcsin (\frac{2}{2} x))) : 2 (arcsin (\frac{x}{2}) + \frac{1}{2} sin (2 arcsin (\frac{x}{2})))

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

Add a constant to the solution

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

x \to 2.99 lim (x^{2} + 3 x)

e^{8}

x \to - 4 lim (x^{2} - 1)

\frac{\partial}{\partial x} (3 x^{2} + y cos (x y))

\frac{\partial}{\partial x} (sin (2 x) cos (8 y))

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