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

The integral of 1/(sqrt(1-9x^2)) is 1/3 arcsin(3x)+C

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

= \int \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = arcsin (3 x)

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

Simplify

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

Add a constant to the solution

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

\frac{d y}{d x} = e^{3 x + 2 y}

n \to \infty lim (\frac{n}{6} sin (\frac{6}{n}))

\int_{0}^{1} (x + 1) e^{x^{2} + 2 x} d x

\int e^{t a n (5 x)} sec^{2} (5 x) d x

\frac{d y}{d x} = - \frac{( x + y ) ^{2}}{2 x y + x ^{2} - 1}

What is the integral of 1/(sqrt(1-9x^2)) ?
The integral of 1/(sqrt(1-9x^2)) is 1/3 arcsin(3x)+C