What is the integral of sqrt(4-4(36x^2)) ?

The integral of sqrt(4-4(36x^2)) is 1/6 (arcsin(6x)+1/2 sin(2arcsin(6x)))+C

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integral of sqrt(4-4(36x^2))

\int 4 - 4 (36 x^{2}) d x

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

Solution steps

\int 4 - 4 \cdot 36 x^{2} d x

Simplify

4 - 4 \cdot 36 x^{2} : 2 1 - 36 x^{2}

= \int 2 1 - 36 x^{2} d x

Take the constant out:

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

= 2 \cdot \int 1 - 36 x^{2} d x

Apply Trigonometric Substitution

: \int \frac{1}{6} cos^{2} (u) d u

= 2 \cdot \int \frac{1}{6} cos^{2} (u) d u

Take the constant out:

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

= 2 \cdot \frac{1}{6} \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \frac{1}{6} \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

= 2 \cdot \frac{1}{6} \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

= 2 \cdot \frac{1}{6} \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)

= 2 \cdot \frac{1}{6} \cdot \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (6 x)

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of sqrt(4-4(36x^2)) ?
The integral of sqrt(4-4(36x^2)) is 1/6 (arcsin(6x)+1/2 sin(2arcsin(6x)))+C