What is the integral of x(1-x^4)^{3/2} ?

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

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integral of x(1-x^4)^{3/2}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{2} \cdot \int (1 - u^{2})^{\frac{3}{2}} d u

Apply Integration By Parts

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

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

\int - 3 u^{2} (1 - u^{2})^{\frac{1}{2}} d u = - \frac{3}{8} (arcsin (u) - \frac{1}{4} sin (4 arcsin (u)))

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

Substitute back

u = x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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