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

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

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

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

- arcsin (2 x) + C

Solution steps

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

Take the constant out:

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

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

Simplify

\frac{1}{\frac{1}{4} - x ^{2}} : \frac{2}{1 - 4 x ^{2}}

= - \int \frac{2}{1 - 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}{1 - 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 (2 x)

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

Simplify

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

= - arcsin (2 x)

Add a constant to the solution

= - arcsin (2 x) + C

(e^{t} (5 - 3 e^{- 2 t}))^{^{'}}

\int 12 x^{3} d x

250

x \to 0 lim (x + 1)

x \to 0 lim (\frac{( e ^{x} - 1 )}{x})

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