What is the integral of 1/(z^2sqrt(9-z^2)) ?

The integral of 1/(z^2sqrt(9-z^2)) is -(sqrt(9-z^2))/(9z)+C

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integral of 1/(z^2sqrt(9-z^2))

\int \frac{1}{z ^{2} 9 - z ^{2}} d z

- \frac{9 - z ^{2}}{9 z} + C

Solution steps

\int \frac{1}{z ^{2} 9 - z ^{2}} d z

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{9 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{9 cos ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

\frac{cot ( u )}{cos ( u ) sin ( u )} : csc^{2} (u)

= \frac{1}{9} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{9} (- cot (u))

Substitute back

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

= \frac{1}{9} (- cot (arcsin (\frac{1}{3} z)))

Simplify

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

= - \frac{9 - z ^{2}}{9 z}

Add a constant to the solution

= - \frac{9 - z ^{2}}{9 z} + C

\frac{4 s + 2}{s ^{2} + 8 s + 16}

\int \frac{1}{x ^{5} + 2 x ^{3} + x} d x

\int_{0}^{10} 400 - 20 x d x

f (t) = 5 t

\int 4 - y d y

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