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

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