What is the integral of (4-x^2)^{0.5} ?

The integral of (4-x^2)^{0.5} is 2arcsin(1/2 x)+sin(2arcsin(1/2 x))+C

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integral of (4-x^2)^{0.5}

\int (4 - x^{2})^{0.5} d x

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

Solution steps

\int (4 - x^{2})^{0.5} d x

Simplify

x^{2} : xx

= \int (4 - xx)^{0.5} d x

xx = x^{2}

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

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} x)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (4-x^2)^{0.5} ?
The integral of (4-x^2)^{0.5} is 2arcsin(1/2 x)+sin(2arcsin(1/2 x))+C