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

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

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integral of 1/((4-x^2)^{1/2)}

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

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

\int 3 x^{8} d x

\frac{d y}{d x} = 0.3 \cdot y

\int \frac{1}{s ( 4 s ) ^{2} - 9} d s

\frac{\partial}{\partial x} (\frac{3 r + s}{t})

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