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

The integral of (sqrt(9-t^2))/(2t^2) is 1/2 (-arcsin(1/3 t)-(sqrt(9-t^2))/t)+C

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

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

\frac{1}{2} (- arcsin (\frac{1}{3} t) - \frac{9 - t ^{2}}{t}) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Trigonometric Substitution

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

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

Rewrite using trig identities

= \frac{1}{2} \cdot \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

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

\int 1 d u = u

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

= \frac{1}{2} (- u - cot (u))

Substitute back

u = arcsin (\frac{1}{3} t)

= \frac{1}{2} (- arcsin (\frac{1}{3} t) - cot (arcsin (\frac{1}{3} t)))

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} (- arcsin (\frac{1}{3} t) - \frac{9 - t ^{2}}{t}) + C

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What is the integral of (sqrt(9-t^2))/(2t^2) ?
The integral of (sqrt(9-t^2))/(2t^2) is 1/2 (-arcsin(1/3 t)-(sqrt(9-t^2))/t)+C