What is the integral of z^3(a^2-z^2)^{3/2} ?

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

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integral of z^3(a^2-z^2)^{3/2}

\int z^{3} (a^{2} - z^{2})^{\frac{3}{2}} d z

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

Solution steps

\int z^{3} (a^{2} - z^{2})^{\frac{3}{2}} d z

a^{2} - z^{2} = a^{2} (1 - \frac{z ^{2}}{a ^{2}})

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

Apply exponent rule:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{z ^{2}}{a ^{2}} = (\frac{z}{a ^{2}})^{2}

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

(a^{2} (1 - (\frac{z}{a ^{2}})^{2}))^{\frac{3}{2}} = (a^{2})^{\frac{3}{2}} ((1 - (\frac{z}{a ^{2}})^{2}))^{\frac{3}{2}}

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

Take the constant out:

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

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= (a^{2})^{\frac{3}{2}} a^{4} \cdot \int - v^{4} (1 - v^{2}) d v

Expand

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

= (a^{2})^{\frac{3}{2}} a^{4} \cdot \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

= (a^{2})^{\frac{3}{2}} a^{4} (- \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}

= (a^{2})^{\frac{3}{2}} a^{4} (- \frac{v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= (a^{2})^{\frac{3}{2}} a^{4} (- \frac{cos ^{5} ( arcsin ( \frac{1}{a} z ) )}{5} + \frac{cos ^{7} ( arcsin ( \frac{1}{a} z ) )}{7})

Simplify

(a^{2})^{\frac{3}{2}} a^{4} (- \frac{cos ^{5} ( arcsin ( \frac{1}{a} z ) )}{5} + \frac{cos ^{7} ( arcsin ( \frac{1}{a} z ) )}{7}) : - \frac{a ^{2} ( - z ^{2} + a ^{2} ) ^{\frac{5}{2}}}{5} + \frac{1}{7} (a^{2} - z^{2})^{\frac{7}{2}}

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

Add a constant to the solution

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

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