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

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

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integral of (u^2)/(2sqrt(1-u^2))

\int \frac{u ^{2}}{2 1 - u ^{2}} d u

\frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))) + C

Solution steps

\int \frac{u ^{2}}{2 1 - u ^{2}} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u ^{2}}{1 - u ^{2}} d u

Apply Trigonometric Substitution

: \int sin^{2} (v) d v

= \frac{1}{2} \cdot \int sin^{2} (v) d v

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{1 - cos ( 2 v )}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 - cos (2 v) d v

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}{2} \cdot \frac{1}{2} (\int 1 d v - \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

= \frac{1}{2} \cdot \frac{1}{2} (v - \frac{1}{2} sin (2 v))

Substitute back

v = arcsin (u)

= \frac{1}{2} \cdot \frac{1}{2} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

Simplify

\frac{1}{2} \cdot \frac{1}{2} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))) : \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

= \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u)))

Add a constant to the solution

= \frac{1}{4} (arcsin (u) - \frac{1}{2} sin (2 arcsin (u))) + C

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