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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

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

\int 1 d u = u

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

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

Substitute back

u = arcsin (t)

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

Simplify

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

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

Add a constant to the solution

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

t^{2} y^{^{'}} + 4 t y - y^{3} = 0, y (1) = \frac{3}{2}

f (x) = x 2^{x}

\frac{d}{d y} (y^{- 3})

\int_{0}^{3} (2 sin (x) - e^{x}) d x

n = 1 \sum \infty \frac{1}{n} \cdot (- 1)^{n}

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