What is the integral of 1/(t^2sqrt(169-t^2)) ?

The integral of 1/(t^2sqrt(169-t^2)) is -(sqrt(169-t^2))/(169t)+C

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integral of 1/(t^2sqrt(169-t^2))

\int \frac{1}{t ^{2} 169 - t ^{2}} d t

- \frac{169 - t ^{2}}{169 t} + C

Solution steps

\int \frac{1}{t ^{2} 169 - t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{169 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{169 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}{169} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{169} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{169} (- cot (u))

Substitute back

u = arcsin (\frac{1}{13} t)

= \frac{1}{169} (- cot (arcsin (\frac{1}{13} t)))

Simplify

\frac{1}{169} (- cot (arcsin (\frac{1}{13} t))) : - \frac{169 - t ^{2}}{169 t}

= - \frac{169 - t ^{2}}{169 t}

Add a constant to the solution

= - \frac{169 - t ^{2}}{169 t} + C

y = \frac{x}{x + 6}, (4, 0.2)

\int_{1}^{e} ln (12 x) d x

y = x^{2}, y = x^{3}

x y^{^{'}} + y = - 8 x y^{2}

f (x) = x^{2} sin (x)

What is the integral of 1/(t^2sqrt(169-t^2)) ?
The integral of 1/(t^2sqrt(169-t^2)) is -(sqrt(169-t^2))/(169t)+C