What is the integral of 1/(sqrt(36-4x^2)) ?

The integral of 1/(sqrt(36-4x^2)) is 1/2 arcsin(x/3)+C

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integral of 1/(sqrt(36-4x^2))

\int \frac{1}{36 - 4 x ^{2}} d x

\frac{1}{2} arcsin (\frac{x}{3}) + C

Solution steps

\int \frac{1}{36 - 4 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= \int \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

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

Substitute back

u = arcsin (\frac{2}{6} x)

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

Simplify

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

= \frac{1}{2} arcsin (\frac{x}{3})

Add a constant to the solution

= \frac{1}{2} arcsin (\frac{x}{3}) + C

\int (4 - 2 x) d x

\frac{d}{d x} (sin (x) + x)

\int \frac{x ^{4} - 2 x + 1}{x ^{2}} d x

\frac{\partial}{\partial x} (sin (x) yz)

5 sin (x - 0.64)

What is the integral of 1/(sqrt(36-4x^2)) ?
The integral of 1/(sqrt(36-4x^2)) is 1/2 arcsin(x/3)+C