What is the integral of sqrt(36-9x^2) ?

The integral of sqrt(36-9x^2) is 6(arcsin(x/2)+1/2 sin(2arcsin(x/2)))+C

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integral of sqrt(36-9x^2)

\int 36 - 9 x^{2} d x

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

Solution steps

\int 36 - 9 x^{2} d x

Apply Trigonometric Substitution

: \int 12 cos^{2} (u) d u

= \int 12 cos^{2} (u) d u

Take the constant out:

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

= 12 \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 12 \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

= 12 \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

= 12 \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)

= 12 \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (\frac{3}{6} x)

= 12 \cdot \frac{1}{2} (arcsin (\frac{3}{6} x) + \frac{1}{2} sin (2 arcsin (\frac{3}{6} x)))

Simplify

12 \cdot \frac{1}{2} (arcsin (\frac{3}{6} x) + \frac{1}{2} sin (2 arcsin (\frac{3}{6} x))) : 6 (arcsin (\frac{x}{2}) + \frac{1}{2} sin (2 arcsin (\frac{x}{2})))

= 6 (arcsin (\frac{x}{2}) + \frac{1}{2} sin (2 arcsin (\frac{x}{2})))

Add a constant to the solution

= 6 (arcsin (\frac{x}{2}) + \frac{1}{2} sin (2 arcsin (\frac{x}{2}))) + C

\frac{d}{d s} (\frac{1 - s ^{2}}{( s ^{2} + 1 ) ^{2}})

\int \frac{1}{3 ( x - 2 )} d x

x^{^{''}} + 2 x^{^{'}} + x = 0, x (0) = 2, x^{^{'}} (0) = 1

\frac{\partial}{\partial y} (4 - y^{2})

sin (t) sin (3 t)

What is the integral of sqrt(36-9x^2) ?
The integral of sqrt(36-9x^2) is 6(arcsin(x/2)+1/2 sin(2arcsin(x/2)))+C