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

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

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integral of sqrt(16-9x^2)

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{16}{3} cos^{2} (u) d u

= \int \frac{16}{3} cos^{2} (u) d u

Take the constant out:

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

= \frac{16}{3} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= \frac{16}{3} \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{16}{3} \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{16}{3} \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{16}{3} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

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

= \frac{16}{3} \cdot \frac{1}{2} (arcsin (\frac{3}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{3}{4} x)))

Simplify

\frac{16}{3} \cdot \frac{1}{2} (arcsin (\frac{3}{4} x) + \frac{1}{2} sin (2 arcsin (\frac{3}{4} x))) : \frac{4}{3} (2 arcsin (\frac{3}{4} x) + sin (2 arcsin (\frac{3}{4} x)))

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (3 x + y^{2})

\frac{d}{d x} (\frac{x}{a ^{2} - x ^{2}})

\frac{d}{d x} (x^{3} - 3 x^{2} + 3 x)

f (x) = x + e^{- 2 x}

6 sin (t)

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