What is the integral of 1/(x^2(1-x^2)^{3/2)} ?

The integral of 1/(x^2(1-x^2)^{3/2)} is -2cot(2arcsin(x))+C

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integral of 1/(x^2(1-x^2)^{3/2)}

\int \frac{1}{x ^{2} ( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

- 2 cot (2 arcsin (x)) + C

Solution steps

\int \frac{1}{x ^{2} ( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{cos ^{3} ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{cos ^{3} ( u ) sin ( u )} d u

Rewrite using trig identities

= \int \frac{8}{1 - cos ( 4 u )} d u

Take the constant out:

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

= 8 \cdot \int \frac{1}{1 - cos ( 4 u )} d u

Apply u-substitution

: \int \frac{1}{4 ( 1 - cos ( v ) )} d v

= 8 \cdot \int \frac{1}{4 ( 1 - cos ( v ) )} d v

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{1 - cos ( v )} d v

Apply u-substitution

: \int \frac{1}{w ^{2}} d w

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{w ^{2}} d w

Apply the Power Rule

: - \frac{1}{w}

= 8 \cdot \frac{1}{4} (- \frac{1}{w})

Substitute back

= 8 \cdot \frac{1}{4} - \frac{1}{tan ( \frac{4 a r c s i n ( x )}{2} )}

Simplify

8 \cdot \frac{1}{4} - \frac{1}{tan ( \frac{4 a r c s i n ( x )}{2} )} : - 2 cot (2 arcsin (x))

= - 2 cot (2 arcsin (x))

Add a constant to the solution

= - 2 cot (2 arcsin (x)) + C

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What is the integral of 1/(x^2(1-x^2)^{3/2)} ?
The integral of 1/(x^2(1-x^2)^{3/2)} is -2cot(2arcsin(x))+C