What is the integral of 1/((1-x^2)^{0.5)} ?

The integral of 1/((1-x^2)^{0.5)} is arcsin(x)+C

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integral of 1/((1-x^2)^{0.5)}

\int \frac{1}{( 1 - x ^{2} ) ^{0.5}} d x

arcsin (x) + C

Solution steps

\int \frac{1}{( 1 - x ^{2} ) ^{0.5}} d x

Simplify

x^{2} : xx

= \int \frac{1}{( 1 - xx ) ^{0.5}} d x

xx = x^{2}

= \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

\int \frac{x ^{2} - 3}{x ^{8}} d x

\int 2 x^{2} (- x^{3} + 3)^{5} d x

\frac{\partial}{\partial t} (3 arcsin (t))

\frac{d y}{d x} = - (1 + y)^{4}

\frac{d}{d x} (\frac{1}{2} x + 4)

What is the integral of 1/((1-x^2)^{0.5)} ?
The integral of 1/((1-x^2)^{0.5)} is arcsin(x)+C