What is the integral of 1/(sqrt(9-36x^2)) ?

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

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integral of 1/(sqrt(9-36x^2))

\int \frac{1}{9 - 36 x ^{2}} d x

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

Solution steps

\int \frac{1}{9 - 36 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{6} d u

= \int \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{4 x}{x + 2})

f (x) = 64 - x^{2}

\frac{\partial}{\partial x} (x y cos (5 yz))

\int (x e^{x^{2}}) d x

\frac{d}{d x} (ln (x^{2} + 12))

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