What is the integral of sqrt(144-9x^2) ?

The integral of sqrt(144-9x^2) is 24(arcsin(x/4)+1/2 sin(2arcsin(x/4)))+C

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integral of sqrt(144-9x^2)

\int 144 - 9 x^{2} d x

24 (arcsin (\frac{x}{4}) + \frac{1}{2} sin (2 arcsin (\frac{x}{4}))) + C

Solution steps

\int 144 - 9 x^{2} d x

Apply Trigonometric Substitution

: \int 48 cos^{2} (u) d u

= \int 48 cos^{2} (u) d u

Take the constant out:

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

= 48 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 48 \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

= 48 \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

= 48 \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)

= 48 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{3}{12} x)

= 48 \cdot \frac{1}{2} (arcsin (\frac{3}{12} x) + \frac{1}{2} sin (2 arcsin (\frac{3}{12} x)))

Simplify

48 \cdot \frac{1}{2} (arcsin (\frac{3}{12} x) + \frac{1}{2} sin (2 arcsin (\frac{3}{12} x))) : 24 (arcsin (\frac{x}{4}) + \frac{1}{2} sin (2 arcsin (\frac{x}{4})))

= 24 (arcsin (\frac{x}{4}) + \frac{1}{2} sin (2 arcsin (\frac{x}{4})))

Add a constant to the solution

= 24 (arcsin (\frac{x}{4}) + \frac{1}{2} sin (2 arcsin (\frac{x}{4}))) + C

x \to - 5 + lim (\frac{3}{x ^{2} - 25})

\frac{d}{d x} (x^{s i n (3 x)})

\frac{\partial}{\partial u} (cos (u + v))

x \to - 4 + lim (\frac{1}{( x + 4 ) ^{2}})

\int \frac{arctan ( 2 x )}{1 + 4 x ^{2}} d x

What is the integral of sqrt(144-9x^2) ?
The integral of sqrt(144-9x^2) is 24(arcsin(x/4)+1/2 sin(2arcsin(x/4)))+C