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

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

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

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

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

Solution steps

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

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 (y)

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

Add a constant to the solution

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

\int (x + 1) (x^{2} + 2 x)^{3} d x

\int \frac{x ^{5} - 5 x}{x ^{4}} d x

- sin (x)

\int sec^{2} (10 x) d x

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