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

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

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

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

arcsin (x) + C

Solution steps

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

Apply exponent rule:

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

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

= \int (1 - 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 (x)

= 1 \cdot arcsin (x)

Simplify

= arcsin (x)

Add a constant to the solution

= arcsin (x) + C

\frac{\partial}{\partial x} (e^{- \frac{x}{y}} + ln (\frac{y}{x}))

\int \frac{4 x}{2 x ^{2} + 3} d x

\int \frac{7}{x ^{2} + 4} d x

t^{2} y^{^{'}} + y = sin (t)

\frac{d}{d x} (- arcsin (\frac{a ^{2} - x ^{2}}{a}))

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