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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{3} \frac{1}{( x - 3 ) ^{2}} d x

\int 7 v (v^{2} + 4)^{2} d v

\int 2 + 2 sin (x) d x

\int 2 + \frac{1}{x} d x

f (x) = \frac{- 8 x}{x ^{2} + 1}, at x = 1

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