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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = 4 - y^{2}

= - \frac{1}{2} \cdot \frac{2}{5} (4 - y^{2})^{\frac{5}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{5} (4 - y^{2})^{\frac{5}{2}} : - \frac{1}{5} (4 - y^{2})^{\frac{5}{2}}

= - \frac{1}{5} (4 - y^{2})^{\frac{5}{2}}

Add a constant to the solution

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

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