What is the integral of x^2sqrt(1-x^2) ?

The integral of x^2sqrt(1-x^2) is 1/8 (arcsin(x)-1/4 sin(4arcsin(x)))+C

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integral of x^2sqrt(1-x^2)

\int x^{2} 1 - x^{2} d x

\frac{1}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))) + C

Solution steps

\int x^{2} 1 - x^{2} d x

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int 1 - cos (4 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

= \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Substitute back

u = arcsin (x)

= \frac{1}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x)))

Add a constant to the solution

= \frac{1}{8} (arcsin (x) - \frac{1}{4} sin (4 arcsin (x))) + C

\int sin (2 x) 5 + sin^{2} (x) d x

\frac{d}{d x} (a e^{- 5 x} + b e^{- 3 x} + c x e^{- x})

\int_{0}^{2} 4.0553 e^{1.8 t} d t

\frac{d}{d x} ((4 x)^{4 x})

\frac{\partial}{\partial x} (\frac{( x + y )}{x + z})

What is the integral of x^2sqrt(1-x^2) ?
The integral of x^2sqrt(1-x^2) is 1/8 (arcsin(x)-1/4 sin(4arcsin(x)))+C