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

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

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integral of sqrt(16-9u^2)

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

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} (v) d v

Rewrite using trig identities

= \frac{16}{3} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

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 v) d v

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 v + \int cos (2 v) d v)

\int 1 d v = v

\int cos (2 v) d v = \frac{1}{2} sin (2 v)

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

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