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

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

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integral of sqrt(4-t^2)

\int 4 - t^{2} d t

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

Solution steps

\int 4 - t^{2} d t

Apply Trigonometric Substitution

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

= \int 4 cos^{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 cos^{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

\frac{d}{d y} (4 y - 1)

x \to 3 - lim (x^{2} - 5 x + 6)

f (x) = x^{4} - 20 x^{2} + 64

\frac{d}{d y} (sin (x y))

\int \frac{6 x + 5}{( 6 x ^{2} + 10 x ) ^{5}} d x

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