What is the integral of 1/(t^2sqrt(100-t^2)) ?

The integral of 1/(t^2sqrt(100-t^2)) is -(sqrt(100-t^2))/(100t)+C

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integral of 1/(t^2sqrt(100-t^2))

\int \frac{1}{t ^{2} 100 - t ^{2}} d t

- \frac{100 - t ^{2}}{100 t} + C

Solution steps

\int \frac{1}{t ^{2} 100 - t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{100 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{100 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}{100} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{100} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{100} (- cot (u))

Substitute back

u = arcsin (\frac{1}{10} t)

= \frac{1}{100} (- cot (arcsin (\frac{1}{10} t)))

Simplify

\frac{1}{100} (- cot (arcsin (\frac{1}{10} t))) : - \frac{100 - t ^{2}}{100 t}

= - \frac{100 - t ^{2}}{100 t}

Add a constant to the solution

= - \frac{100 - t ^{2}}{100 t} + C

x \to 0 lim (ln (cos (x)))

4 t y^{^{'}} + 4 = 5 + t

x \to 0 lim (0^{0})

f (x) = \frac{x ^{2}}{x + 3}

\frac{\partial}{\partial x} (4 x y + 3 yz + 5 x y - 48)

What is the integral of 1/(t^2sqrt(100-t^2)) ?
The integral of 1/(t^2sqrt(100-t^2)) is -(sqrt(100-t^2))/(100t)+C