What is the integral of-sqrt(16-x^2) ?

The integral of-sqrt(16-x^2) is -8(arcsin(1/4 x)+1/2 sin(2arcsin(1/4 x)))+C

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integral of-sqrt(16-x^2)

\int - 16 - x^{2} d x

- 8 (arcsin (\frac{1}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} x))) + C

Solution steps

\int - 16 - x^{2} d x

Take the constant out:

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

= - \int 16 - x^{2} d x

Apply Trigonometric Substitution

: \int 16 cos^{2} (u) d u

= - \int 16 cos^{2} (u) d u

Take the constant out:

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

= - 16 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= - 16 \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

= - 16 \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

= - 16 \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)

= - 16 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{1}{4} x)

= - 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} x)))

Simplify

- 16 \cdot \frac{1}{2} (arcsin (\frac{1}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} x))) : - 8 (arcsin (\frac{1}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} x)))

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

Add a constant to the solution

= - 8 (arcsin (\frac{1}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{1}{4} x))) + C

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What is the integral of-sqrt(16-x^2) ?
The integral of-sqrt(16-x^2) is -8(arcsin(1/4 x)+1/2 sin(2arcsin(1/4 x)))+C