What is the integral of x^5sqrt((1-x^2)) ?

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

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integral of x^5sqrt((1-x^2))

\int x^{5} (1 - x^{2}) d x

- \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} + \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} - \frac{1}{7} (1 - x^{2})^{\frac{7}{2}} + C

Solution steps

\int x^{5} 1 - x^{2} d x

Apply Trigonometric Substitution

: \int sin^{5} (u) cos^{2} (u) d u

= \int sin^{5} (u) cos^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

= \int - v^{2} + 2 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^{2} d v + \int 2 v^{4} d v - \int v^{6} d v

\int v^{2} d v = \frac{v ^{3}}{3}

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

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

= - \frac{v ^{3}}{3} + \frac{2 v ^{5}}{5} - \frac{v ^{7}}{7}

Substitute back

= - \frac{cos ^{3} ( arcsin ( x ) )}{3} + \frac{2 cos ^{5} ( arcsin ( x ) )}{5} - \frac{cos ^{7} ( arcsin ( x ) )}{7}

Simplify

- \frac{cos ^{3} ( arcsin ( x ) )}{3} + \frac{2 cos ^{5} ( arcsin ( x ) )}{5} - \frac{cos ^{7} ( arcsin ( x ) )}{7} : - \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} + \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} - \frac{1}{7} (1 - x^{2})^{\frac{7}{2}}

= - \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} + \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} - \frac{1}{7} (1 - x^{2})^{\frac{7}{2}}

Add a constant to the solution

= - \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} + \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} - \frac{1}{7} (1 - x^{2})^{\frac{7}{2}} + C

\int cot^{5} (x) csc^{6} (x) d x

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

\int - 3^{4} x d x

\int \frac{x ^{7}}{( 1 + x ^{4} ) ^{\frac{3}{2}}} d x

\int \frac{x ^{2} + 16}{x ^{6}} d x

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