What is the integral of t/(sqrt(8-t^2)) ?

The integral of t/(sqrt(8-t^2)) is -sqrt(8-t^2)+C

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integral of t/(sqrt(8-t^2))

\int \frac{t}{8 - t ^{2}} d t

- 8 - t^{2} + C

Solution steps

\int \frac{t}{8 - t ^{2}} d t

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{2 u} 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 \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 8 - t^{2}

= - \frac{1}{2} \cdot 2 (8 - t^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{2} \cdot 2 (8 - t^{2})^{\frac{1}{2}} : - 8 - t^{2}

= - 8 - t^{2}

Add a constant to the solution

= - 8 - t^{2} + C

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What is the integral of t/(sqrt(8-t^2)) ?
The integral of t/(sqrt(8-t^2)) is -sqrt(8-t^2)+C