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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

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

\int e^{- s t} t d t

f (x) = 2 x^{3} - x^{2} + 9, at x = 2

x \to \infty lim (- \frac{π}{2})

x \to 0 lim ((1 - \frac{1}{x})^{x^{2}})

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