What is the integral of 1/(sqrt(4-9x^2)) ?

The integral of 1/(sqrt(4-9x^2)) is 1/3 arcsin(3/2 x)+C

integral of 1/(sqrt(4-9x^2))

\int \frac{1}{4 - 9 x ^{2}} d x

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

Solution steps

\int \frac{1}{4 - 9 x ^{2}} d x

Apply Trigonometric Substitution

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

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

Simplify

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

Add a constant to the solution

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

x \to 3 lim (6 - 2 x)

x \to i lim (x^{2} + 2 x)

\int \frac{1 - x}{x} d x

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

\frac{d}{d x} (x + ln x^{2} - 4 x + 3)

What is the integral of 1/(sqrt(4-9x^2)) ?
The integral of 1/(sqrt(4-9x^2)) is 1/3 arcsin(3/2 x)+C