What is the integral of 3tsqrt(16-t^2) ?

The integral of 3tsqrt(16-t^2) is -(16-t^2)^{3/2}+C

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integral of 3tsqrt(16-t^2)

\int 3 t 16 - t^{2} d t

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

Solution steps

\int 3 t 16 - t^{2} d t

Take the constant out:

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

= 3 \cdot \int t 16 - t^{2} d t

Apply u-substitution

: \int - \frac{u}{2} d u

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

Take the constant out:

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

= 3 (- \frac{1}{2} \cdot \int u d u)

Apply the Power Rule

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

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

Substitute back

u = 16 - t^{2}

= 3 (- \frac{1}{2} \cdot \frac{2}{3} (16 - t^{2})^{\frac{3}{2}})

Simplify

3 (- \frac{1}{2} \cdot \frac{2}{3} (16 - t^{2})^{\frac{3}{2}}) : - (16 - t^{2})^{\frac{3}{2}}

= - (16 - t^{2})^{\frac{3}{2}}

Add a constant to the solution

= - (16 - t^{2})^{\frac{3}{2}} + C

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What is the integral of 3tsqrt(16-t^2) ?
The integral of 3tsqrt(16-t^2) is -(16-t^2)^{3/2}+C