What is the integral of (1/2)/(sqrt(4-3x^2)) ?

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

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integral of (1/2)/(sqrt(4-3x^2))

\int \frac{\frac{1}{2}}{4 - 3 x ^{2}} d x

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

Solution steps

\int \frac{\frac{1}{2}}{4 - 3 x ^{2}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{4 - 3 x ^{2}} d x

Apply Trigonometric Substitution

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

= \frac{1}{2} \cdot \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}{2} \cdot \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot \frac{1}{3} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int arcsec (y) d y

\int sec^{2} (θ) tan^{6} (θ) d θ

e^{p} (p + p p)

\frac{\partial}{\partial x} (\frac{y}{1 + x ^{2} y ^{2}})

y^{^{'}} - a \cdot y + b = 0

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