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

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

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integral of (x^2)/(sqrt(2-x^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 2 sin^{2} (u) d u

= \int 2 sin^{2} (u) d u

Take the constant out:

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

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

Rewrite using trig identities

= 2 \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

= 2 \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

= 2 \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)

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

Substitute back

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

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

Simplify

2 \cdot \frac{1}{2} (arcsin (\frac{1}{2} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))) : arcsin (\frac{1}{2} x) - \frac{1}{2} sin (2 arcsin (\frac{1}{2} x))

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x e^{3 y})

\frac{\partial}{\partial x} (\frac{4 x - 3 y}{4 x + 3 y})

f (x) = - 16 t^{2} - 8 t + 64

(x^{2} - 3) (x^{3} - 2 x)

\int_{0}^{\frac{π}{4}} (cos (2 x)) d x

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