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

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

(x)^{2} y^{^{''}} + 3 (x) y^{^{'}} - 3 y = (x)^{3}

d er i v a t i v e \frac{1}{t ^{2}}

\int_{0}^{1} x \cdot e^{x} d x

d er i v a t i v e y = x - x^{3}

\int \frac{x ^{2} - 5}{( x + 1 ) ^{3}} 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