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

The integral of (sqrt(16-t^2))/(t^2) is -arcsin(1/4 t)-(sqrt(16-t^2))/t+C

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

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

- arcsin (\frac{1}{4} t) - \frac{16 - t ^{2}}{t} + C

Solution steps

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

Apply Trigonometric Substitution

: \int cot^{2} (u) d u

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

Rewrite using trig identities

= \int - 1 + csc^{2} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int 1 d u + \int csc^{2} (u) d u

\int 1 d u = u

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

= - u - cot (u)

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

= - arcsin (\frac{1}{4} t) - \frac{16 - t ^{2}}{t} + C

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What is the integral of (sqrt(16-t^2))/(t^2) ?
The integral of (sqrt(16-t^2))/(t^2) is -arcsin(1/4 t)-(sqrt(16-t^2))/t+C