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

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

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integral of sqrt(9-16x^2)

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

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

Solution steps

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

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} x)

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

Simplify

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

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

Add a constant to the solution

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

\int (\frac{1}{x ^{2} - 9}) d x

x, - x + 6, 0

\frac{\partial}{\partial x} ((x^{2} + 9 x - 4 y)^{6})

\frac{d}{d x} (\frac{2 x - 1}{x})

\int \frac{1}{- cos ( u ) + 1} d u

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