What is the integral of (6/(sqrt(4-x^2))) ?

The integral of (6/(sqrt(4-x^2))) is 6arcsin(1/2 x)+C

Popular >

integral of (6/(sqrt(4-x^2)))

\int (\frac{6}{4 - x ^{2}}) d x

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

Solution steps

\int \frac{6}{4 - x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{4 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 6 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 6 \cdot 1 \cdot u

Substitute back

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

= 6 \cdot 1 \cdot arcsin (\frac{1}{2} x)

Simplify

= 6 arcsin (\frac{1}{2} x)

Add a constant to the solution

= 6 arcsin (\frac{1}{2} x) + C

\frac{d}{d y} (e^{x^{2} + y^{2}} sin (x y))

x^{4} - 3 x

x \to - 0.1 lim (\frac{e ^{x} - 1}{x})

\frac{\partial}{\partial x} (\frac{3}{4 y ^{4} x ^{\frac{1}{4}}})

\frac{d x}{d t} = 7 x t^{4}

What is the integral of (6/(sqrt(4-x^2))) ?
The integral of (6/(sqrt(4-x^2))) is 6arcsin(1/2 x)+C