What is the integral of y^2((1-y^2)^{1/2}) ?

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

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integral of y^2((1-y^2)^{1/2})

integral

y^{2} ((1 - y^{2})^{\frac{1}{2}})

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

Solution steps

\int y^{2} (1 - y^{2})^{\frac{1}{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

x^{4} + 2 x + 3

\frac{( e ^{x} )}{x ^{- 1}}

2 y lo g_{10} (y) + 1

6 x \cdot e^{- 2} x

\frac{t ^{2}}{( 1 + t ^{2} )}

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