What is the integral of w/(sqrt(16-w^2)) ?

The integral of w/(sqrt(16-w^2)) is -sqrt(16-w^2)+C

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integral of w/(sqrt(16-w^2))

\int \frac{w}{16 - w ^{2}} d w

- 16 - w^{2} + C

Solution steps

\int \frac{w}{16 - w ^{2}} d w

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 = 16 - w^{2}

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

Simplify

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

= - 16 - w^{2}

Add a constant to the solution

= - 16 - w^{2} + C

\frac{d}{d x} (\frac{x}{x ^{2} + 7})

x \to 0 lim (7 cot^{4} (3 x))

\int \frac{cos ( x )}{x} d x

f (x) = 5 (7 + x)

\frac{d}{d x} (\frac{1}{( 1 + x ^{2} )})

What is the integral of w/(sqrt(16-w^2)) ?
The integral of w/(sqrt(16-w^2)) is -sqrt(16-w^2)+C