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

The integral of x^3(1-x^2)^{1/4} is -2/45 (5x^2+4)(-x^2+1)^{5/4}+C

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

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

- \frac{2}{45} (5 x^{2} + 4) (- x^{2} + 1)^{\frac{5}{4}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{( - u + 1 ) u ^{\frac{1}{4}}}{2} d u

= \int - \frac{( - u + 1 ) u ^{\frac{1}{4}}}{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 (- u + 1) u^{\frac{1}{4}} d u

Expand

(- u + 1) u^{\frac{1}{4}} : - u^{\frac{5}{4}} + u^{\frac{1}{4}}

= - \frac{1}{2} \cdot \int - u^{\frac{5}{4}} + u^{\frac{1}{4}} 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

= - \frac{1}{2} (- \int u^{\frac{5}{4}} d u + \int u^{\frac{1}{4}} d u)

\int u^{\frac{5}{4}} d u = \frac{4}{9} u^{\frac{9}{4}}

\int u^{\frac{1}{4}} d u = \frac{4}{5} u^{\frac{5}{4}}

= - \frac{1}{2} (- \frac{4}{9} u^{\frac{9}{4}} + \frac{4}{5} u^{\frac{5}{4}})

Substitute back

u = 1 - x^{2}

= - \frac{1}{2} (- \frac{4}{9} (1 - x^{2})^{\frac{9}{4}} + \frac{4}{5} (1 - x^{2})^{\frac{5}{4}})

Simplify

- \frac{1}{2} (- \frac{4}{9} (1 - x^{2})^{\frac{9}{4}} + \frac{4}{5} (1 - x^{2})^{\frac{5}{4}}) : - \frac{2}{45} (5 x^{2} + 4) (- x^{2} + 1)^{\frac{5}{4}}

= - \frac{2}{45} (5 x^{2} + 4) (- x^{2} + 1)^{\frac{5}{4}}

Add a constant to the solution

= - \frac{2}{45} (5 x^{2} + 4) (- x^{2} + 1)^{\frac{5}{4}} + C

\frac{d ^{2}}{d x ^{2}} (6 sec (x))

\int_{0}^{3 π} sin^{2} (x) d x

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

y^{^{''}} - 4 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = - 8

f (x) = \frac{x ^{3}}{9}

What is the integral of x^3(1-x^2)^{1/4} ?
The integral of x^3(1-x^2)^{1/4} is -2/45 (5x^2+4)(-x^2+1)^{5/4}+C