What is the integral of (sqrt(9-w^2))/(w^2) ?

The integral of (sqrt(9-w^2))/(w^2) is -arcsin(1/3 w)-(sqrt(9-w^2))/w+C

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integral of (sqrt(9-w^2))/(w^2)

\int \frac{9 - w ^{2}}{w ^{2}} d w

- arcsin (\frac{1}{3} w) - \frac{9 - w ^{2}}{w} + C

Solution steps

\int \frac{9 - w ^{2}}{w ^{2}} d w

Apply Trigonometric Substitution

: \int cot^{2} (u) d u

= \int cot^{2} (u) d u

Rewrite using trig identities

= \int - 1 + csc^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d u + \int csc^{2} (u) d u

\int 1 d u = u

\int csc^{2} (u) d u = - cot (u)

= - u - cot (u)

Substitute back

u = arcsin (\frac{1}{3} w)

= - arcsin (\frac{1}{3} w) - cot (arcsin (\frac{1}{3} w))

Simplify

- arcsin (\frac{1}{3} w) - cot (arcsin (\frac{1}{3} w)) : - arcsin (\frac{1}{3} w) - \frac{9 - w ^{2}}{w}

= - arcsin (\frac{1}{3} w) - \frac{9 - w ^{2}}{w}

Add a constant to the solution

= - arcsin (\frac{1}{3} w) - \frac{9 - w ^{2}}{w} + C

\frac{d}{d y} (0)

\frac{d}{d x} (tan (x) - e^{- x})

\int_{0}^{1} e^{3 x + 1} d x

\frac{d}{d x} (\frac{( x ^{2} + 4 )}{x})

y = 2 x^{3} - 8 x, (2, 0)

What is the integral of (sqrt(9-w^2))/(w^2) ?
The integral of (sqrt(9-w^2))/(w^2) is -arcsin(1/3 w)-(sqrt(9-w^2))/w+C