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

The integral of (t^2)/(sqrt(4t-t^2)) is 6arcsin(1/2 t-1)-sin(2arcsin(1/2 t-1))-4sqrt(-t^2+4t)+C

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

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

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

Solution steps

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

Complete the square

4 t - t^{2} : - (t - 2)^{2} + 4

= \int \frac{t ^{2}}{- ( t - 2 ) ^{2} + 4} d t

Apply u-substitution

: \int \frac{( u + 2 ) ^{2}}{- u ^{2} + 4} d u

= \int \frac{( u + 2 ) ^{2}}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

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

= \int (2 sin (v) + 2)^{2} d v

Expand

(2 sin (v) + 2)^{2} : 4 sin^{2} (v) + 8 sin (v) + 4

= \int 4 sin^{2} (v) + 8 sin (v) + 4 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

= \int 4 sin^{2} (v) d v + \int 8 sin (v) d v + \int 4 d v

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

\int 8 sin (v) d v = - 8 cos (v)

\int 4 d v = 4 v

= 2 v - sin (2 v) - 8 cos (v) + 4 v

Substitute back

= 2 arcsin (\frac{1}{2} (t - 2)) - sin (2 arcsin (\frac{1}{2} (t - 2))) - 8 cos (arcsin (\frac{1}{2} (t - 2))) + 4 arcsin (\frac{1}{2} (t - 2))

Simplify

2 arcsin (\frac{1}{2} (t - 2)) - sin (2 arcsin (\frac{1}{2} (t - 2))) - 8 cos (arcsin (\frac{1}{2} (t - 2))) + 4 arcsin (\frac{1}{2} (t - 2)) : 6 arcsin (\frac{1}{2} t - 1) - sin (2 arcsin (\frac{1}{2} t - 1)) - 4 - t^{2} + 4 t

= 6 arcsin (\frac{1}{2} t - 1) - sin (2 arcsin (\frac{1}{2} t - 1)) - 4 - t^{2} + 4 t

Add a constant to the solution

= 6 arcsin (\frac{1}{2} t - 1) - sin (2 arcsin (\frac{1}{2} t - 1)) - 4 - t^{2} + 4 t + C

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What is the integral of (t^2)/(sqrt(4t-t^2)) ?
The integral of (t^2)/(sqrt(4t-t^2)) is 6arcsin(1/2 t-1)-sin(2arcsin(1/2 t-1))-4sqrt(-t^2+4t)+C