What is the integral of (4-y^2)^{1/2} ?

The integral of (4-y^2)^{1/2} is 2arcsin(1/2 y)+sin(2arcsin(1/2 y))+C

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integral of (4-y^2)^{1/2}

\int (4 - y^{2})^{\frac{1}{2}} d y

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

Solution steps

\int (4 - y^{2})^{\frac{1}{2}} d y

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} y)

= 4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} y)))

Simplify

4 \cdot \frac{1}{2} (arcsin (\frac{1}{2} y) + \frac{1}{2} sin (2 arcsin (\frac{1}{2} y))) : 2 arcsin (\frac{1}{2} y) + sin (2 arcsin (\frac{1}{2} y))

= 2 arcsin (\frac{1}{2} y) + sin (2 arcsin (\frac{1}{2} y))

Add a constant to the solution

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

\frac{d}{d x} (sin (1 - x) cos^{3} (x))

f (x) = \frac{e ^{x}}{2 + 5 x}

x + \frac{y d y}{d x} = 0

6 y^{^{'}} - 2 y = x y^{4}, y (0) = - 2

f (x) = \frac{6 x + 16}{x + 3}

What is the integral of (4-y^2)^{1/2} ?
The integral of (4-y^2)^{1/2} is 2arcsin(1/2 y)+sin(2arcsin(1/2 y))+C