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

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

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

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

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

Solution steps

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

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} t)

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

Simplify

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

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

Add a constant to the solution

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

\frac{d ^{2}}{d y ^{2}} (cos (x y^{2}))

\int 5 x ln (3 x) d x

f (x) = arctan (\frac{1}{x ^{2}})

f (x) = x (2 - x) (x + 3)

y^{^{'}} = 4 y + y^{4}, y (0) = 1

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