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

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

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

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

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

Solution steps

\int \frac{1}{1 - 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 (6 x)

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

Simplify

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

Add a constant to the solution

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

y^{^{'''}} - 4 y^{^{''}} - y^{^{'}} + 4 y = 0

x \to 2 lim (\frac{3 ^{x} - 9}{x - 2})

\frac{\partial}{\partial x} (9 x e^{x^{2} + y^{2}})

\frac{d}{d x} (\frac{1}{5} x^{5} - 2 x^{3} + 3 x)

x \to - 1 lim (6 x + 3)

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