What is the integral of 1/(sqrt(3-2x^2)) ?

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

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integral of 1/(sqrt(3-2x^2))

\int \frac{1}{3 - 2 x ^{2}} d x

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

Solution steps

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

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

Add a constant to the solution

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

e^{t} cos (4 t)

\int x^{\frac{5}{2}} ln (x) d x

\frac{\partial}{\partial x} ((3 x - 2 x^{2})^{3})

\int (\frac{1 + x}{x})^{2} d x

\frac{d y}{d x} = \frac{1}{y ^{2}}

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