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

The integral of 1/(sqrt(1-x^2))1-x^2) is d 1/2 (arcsin(x)+1/2 sin(2arcsin(x)))+C

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integral of 1/(sqrt(1-x^2))1-x^2)

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

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

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

Apply Trigonometric Substitution

: \int cos^{2} (u) d u

= d \cdot \int cos^{2} (u) d u

Rewrite using trig identities

= d \cdot \int \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= d \frac{1}{2} \cdot \int 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= d \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= d \frac{1}{2} (u + \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (x)

= d \frac{1}{2} (arcsin (x) + \frac{1}{2} sin (2 arcsin (x)))

Add a constant to the solution

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

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What is the integral of 1/(sqrt(1-x^2))1-x^2) ?
The integral of 1/(sqrt(1-x^2))1-x^2) is d 1/2 (arcsin(x)+1/2 sin(2arcsin(x)))+C