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

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

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

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

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

Solution steps

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

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 - x ^{2}}{x ^{2}} d x

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} x)

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

Simplify

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

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

Add a constant to the solution

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

(x^{2} - 2 x + 2) e^{\frac{5 x}{2}}

x^{2} + y^{2} = 26

\frac{d}{d x} (- x^{3} + 5 x^{2} - 8 x)

\int_{0}^{2} x^{3} 7 - x^{2} d x

\frac{d}{d x} (1 - e^{- 0.1 x})

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