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

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

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

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

arcsin (x) + C

Solution steps

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

Factor

1 - x^{2} : - (- 1 + x^{2})

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

Simplify

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

Use the common integral:

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

= arcsin (x)

Add a constant to the solution

= arcsin (x) + C

\int sin (t x) d x

\int_{0}^{l n (x)} sin (e^{t}) d t

\frac{d}{d x} ((3 x^{2} + 4 x + 8) x - 1)

n = 3 \sum \infty \frac{8}{n ^{2} - 1}

4 x^{3} + 6 x^{2}

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