What is the integral of x^4 1/(sqrt(1-x^2)) ?

The integral of x^4 1/(sqrt(1-x^2)) is -1/4 x^3sqrt(1-x^2)+3/8 (arcsin(x)-1/2 sin(2arcsin(x)))+C

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integral of x^4 1/(sqrt(1-x^2))

\int x^{4} \frac{1}{1 - x ^{2}} d x

- \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) + C

Solution steps

\int x^{4} \frac{1}{1 - x ^{2}} d x

Simplify

x^{4} \frac{1}{1 - x ^{2}} : \frac{x ^{4}}{1 - x ^{2}}

= \int \frac{x ^{4}}{1 - x ^{2}} d x

Apply Trigonometric Substitution

= \int sin^{4} (u) d u

Apply Integral Reduction

= - \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u

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

= - \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = arcsin (x)

= - \frac{cos ( arcsin ( x ) ) sin ^{3} ( arcsin ( x ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

Simplify

- \frac{cos ( arcsin ( x ) ) sin ^{3} ( arcsin ( x ) )}{4} + \frac{3}{4} \cdot \frac{1}{2} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) : - \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

= - \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x)))

Add a constant to the solution

= - \frac{1}{4} x^{3} 1 - x^{2} + \frac{3}{8} (arcsin (x) - \frac{1}{2} sin (2 arcsin (x))) + C

What is the integral of x^4 1/(sqrt(1-x^2)) ?
The integral of x^4 1/(sqrt(1-x^2)) is -1/4 x^3sqrt(1-x^2)+3/8 (arcsin(x)-1/2 sin(2arcsin(x)))+C