What is the integral of sqrt(9-16y^2) ?

The integral of sqrt(9-16y^2) is 9/8 (arcsin(4/3 y)+1/2 sin(2arcsin(4/3 y)))+C

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integral of sqrt(9-16y^2)

\int 9 - 16 y^{2} d y

\frac{9}{8} (arcsin (\frac{4}{3} y) + \frac{1}{2} sin (2 arcsin (\frac{4}{3} y))) + C

Solution steps

\int 9 - 16 y^{2} d y

Apply Trigonometric Substitution

: \int \frac{9}{4} cos^{2} (u) d u

= \int \frac{9}{4} cos^{2} (u) d u

Take the constant out:

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

= \frac{9}{4} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= \frac{9}{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

= \frac{9}{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

= \frac{9}{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)

= \frac{9}{4} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{4}{3} y)

= \frac{9}{4} \cdot \frac{1}{2} (arcsin (\frac{4}{3} y) + \frac{1}{2} sin (2 arcsin (\frac{4}{3} y)))

Simplify

\frac{9}{4} \cdot \frac{1}{2} (arcsin (\frac{4}{3} y) + \frac{1}{2} sin (2 arcsin (\frac{4}{3} y))) : \frac{9}{8} (arcsin (\frac{4}{3} y) + \frac{1}{2} sin (2 arcsin (\frac{4}{3} y)))

= \frac{9}{8} (arcsin (\frac{4}{3} y) + \frac{1}{2} sin (2 arcsin (\frac{4}{3} y)))

Add a constant to the solution

= \frac{9}{8} (arcsin (\frac{4}{3} y) + \frac{1}{2} sin (2 arcsin (\frac{4}{3} y))) + C

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What is the integral of sqrt(9-16y^2) ?
The integral of sqrt(9-16y^2) is 9/8 (arcsin(4/3 y)+1/2 sin(2arcsin(4/3 y)))+C