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

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

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

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

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

Solution steps

\int \frac{1}{9 - 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}{3} x)

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

Simplify

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

Add a constant to the solution

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

cos (2 x) \cdot ln (3 x - 5)

\frac{x ^{2}}{4 + x}

\frac{\partial}{\partial x} (6 e^{x^{5} y^{4}})

x \to 0 lim (\frac{x}{tan ( 7 x )})

\frac{\partial}{\partial x} (e^{- x yz})

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