What is the integral of t(sqrt(4-t^2)) ?

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

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integral of t(sqrt(4-t^2))

\int t (4 - t^{2}) d t

- \frac{1}{3} (4 - t^{2})^{\frac{3}{2}} + C

Solution steps

\int t 4 - t^{2} d t

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = 4 - t^{2}

= - \frac{1}{2} \cdot \frac{2}{3} (4 - t^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (4 - t^{2})^{\frac{3}{2}} : - \frac{1}{3} (4 - t^{2})^{\frac{3}{2}}

= - \frac{1}{3} (4 - t^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (4 - t^{2})^{\frac{3}{2}} + C

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