What is the integral of (2-t^2)sqrt(6t-t^3) ?

The integral of (2-t^2)sqrt(6t-t^3) is 2/9 (6t-t^3)^{3/2}+C

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integral of (2-t^2)sqrt(6t-t^3)

\int (2 - t^{2}) 6 t - t^{3} d t

\frac{2}{9} (6 t - t^{3})^{\frac{3}{2}} + C

Solution steps

\int (2 - t^{2}) 6 t - t^{3} d t

Apply u-substitution

: \int \frac{u}{3} d u

= \int \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 6 t - t^{3}

= \frac{1}{3} \cdot \frac{2}{3} (6 t - t^{3})^{\frac{3}{2}}

Simplify

\frac{1}{3} \cdot \frac{2}{3} (6 t - t^{3})^{\frac{3}{2}} : \frac{2}{9} (6 t - t^{3})^{\frac{3}{2}}

= \frac{2}{9} (6 t - t^{3})^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{9} (6 t - t^{3})^{\frac{3}{2}} + C

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What is the integral of (2-t^2)sqrt(6t-t^3) ?
The integral of (2-t^2)sqrt(6t-t^3) is 2/9 (6t-t^3)^{3/2}+C