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

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

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

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

arcsin (y) + C

Solution steps

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

Apply exponent rule:

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

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

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

Apply Trigonometric Substitution

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

u = arcsin (y)

= 1 \cdot arcsin (y)

Simplify

= arcsin (y)

Add a constant to the solution

= arcsin (y) + C

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