What is the integral of (t^2)/(sqrt(4-t^2)) ?

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

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integral of (t^2)/(sqrt(4-t^2))

\int \frac{t ^{2}}{4 - t ^{2}} d t

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

Solution steps

\int \frac{t ^{2}}{4 - t ^{2}} d t

Apply Trigonometric Substitution

: \int 4 sin^{2} (u) d u

= \int 4 sin^{2} (u) d u

Take the constant out:

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

= 4 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 4 \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

= 4 \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

= 4 \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)

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

Substitute back

u = arcsin (\frac{1}{2} t)

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{1} x (x - 1) e^{- x^{2}} d x

\int (5 x^{2} - 5 sin (x)) d x

\frac{d}{d x} (sinh (2 x))

\int 4 cos (4 x) d x

x \to 1 + lim ((x - 1)^{(x - 1)})

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