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

The integral of x^3(1-x^2)^{3/2} is -1/5 (1-x^2)^{5/2}+1/7 (1-x^2)^{7/2}+C

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

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

- \frac{1}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{1}{7} (1 - x^{2})^{\frac{7}{2}} + C

Solution steps

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

Apply Trigonometric Substitution

: \int cos^{4} (u) sin^{3} (u) d u

= \int cos^{4} (u) sin^{3} (u) d u

Rewrite using trig identities

= \int (1 - cos^{2} (u)) sin (u) cos^{4} (u) d u

Apply u-substitution

: \int - v^{4} (1 - v^{2}) d v

= \int - v^{4} (1 - v^{2}) d v

Expand

- v^{4} (1 - v^{2}) : - v^{4} + v^{6}

= \int - v^{4} + v^{6} d v

Apply the Sum Rule:

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

= - \int v^{4} d v + \int v^{6} d v

\int v^{4} d v = \frac{v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= - \frac{v ^{5}}{5} + \frac{v ^{7}}{7}

Substitute back

= - \frac{cos ^{5} ( arcsin ( x ) )}{5} + \frac{cos ^{7} ( arcsin ( x ) )}{7}

Simplify

- \frac{cos ^{5} ( arcsin ( x ) )}{5} + \frac{cos ^{7} ( arcsin ( x ) )}{7} : - \frac{1}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{1}{7} (1 - x^{2})^{\frac{7}{2}}

= - \frac{1}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{1}{7} (1 - x^{2})^{\frac{7}{2}}

Add a constant to the solution

= - \frac{1}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{1}{7} (1 - x^{2})^{\frac{7}{2}} + C

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\int_{0}^{3} sin^{2} (m x) d x

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\frac{d}{d x} (x + \frac{1}{x ^{3}})

What is the integral of x^3(1-x^2)^{3/2} ?
The integral of x^3(1-x^2)^{3/2} is -1/5 (1-x^2)^{5/2}+1/7 (1-x^2)^{7/2}+C