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

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

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integral of sqrt(16-y^2)

\int 16 - y^{2} d y

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

Solution steps

\int 16 - y^{2} d y

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} y)

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

Simplify

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

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

Add a constant to the solution

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

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