What is the integral of 1/(t^2sqrt(16-t^2)) ?

The integral of 1/(t^2sqrt(16-t^2)) is -(sqrt(16-t^2))/(16t)+C

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integral of 1/(t^2sqrt(16-t^2))

\int \frac{1}{t ^{2} 16 - t ^{2}} d t

- \frac{16 - t ^{2}}{16 t} + C

Solution steps

\int \frac{1}{t ^{2} 16 - t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{16 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{16 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}{16} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{16} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{16} (- cot (u))

Substitute back

u = arcsin (\frac{1}{4} t)

= \frac{1}{16} (- cot (arcsin (\frac{1}{4} t)))

Simplify

\frac{1}{16} (- cot (arcsin (\frac{1}{4} t))) : - \frac{16 - t ^{2}}{16 t}

= - \frac{16 - t ^{2}}{16 t}

Add a constant to the solution

= - \frac{16 - t ^{2}}{16 t} + C

y = 2 - x^{3}, (1, 1)

y^{^{'}} = (t + y)^{2} - 1

\int \frac{3 x ^{3} - 4 x ^{2} - 14 x + 8}{x ^{2} - 3 x + 1} d x

f (x) = ∣ x - 9 ∣

arctan (2 t)

What is the integral of 1/(t^2sqrt(16-t^2)) ?
The integral of 1/(t^2sqrt(16-t^2)) is -(sqrt(16-t^2))/(16t)+C