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

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

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

\int x^{2} \cdot 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

y^{^{''}} + 121 y = cos (6 t)

\int \frac{3 ^{x}}{x} d x

3 y^{^{''}} + 8 y^{^{'}} - 3 y = 0

\int (1 + cos^{2} (x)) d x

n = 1 \sum \infty \frac{7 ^{n}}{6 ^{n}}

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