What is the integral of (sqrt(25-x^2))/(15x^2) ?

The integral of (sqrt(25-x^2))/(15x^2) is 1/15 (-arcsin(1/5 x)-(sqrt(25-x^2))/x)+C

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integral of (sqrt(25-x^2))/(15x^2)

\int \frac{25 - x ^{2}}{15 x ^{2}} d x

\frac{1}{15} (- arcsin (\frac{1}{5} x) - \frac{25 - x ^{2}}{x}) + C

Solution steps

\int \frac{25 - x ^{2}}{15 x ^{2}} d x

Take the constant out:

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

= \frac{1}{15} \cdot \int \frac{25 - x ^{2}}{x ^{2}} d x

Apply Trigonometric Substitution

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

= \frac{1}{15} \cdot \int cot^{2} (u) d u

Rewrite using trig identities

= \frac{1}{15} \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}{15} (- \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}{15} (- u - cot (u))

Substitute back

u = arcsin (\frac{1}{5} x)

= \frac{1}{15} (- arcsin (\frac{1}{5} x) - cot (arcsin (\frac{1}{5} x)))

Simplify

\frac{1}{15} (- arcsin (\frac{1}{5} x) - cot (arcsin (\frac{1}{5} x))) : \frac{1}{15} (- arcsin (\frac{1}{5} x) - \frac{25 - x ^{2}}{x})

= \frac{1}{15} (- arcsin (\frac{1}{5} x) - \frac{25 - x ^{2}}{x})

Add a constant to the solution

= \frac{1}{15} (- arcsin (\frac{1}{5} x) - \frac{25 - x ^{2}}{x}) + C

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What is the integral of (sqrt(25-x^2))/(15x^2) ?
The integral of (sqrt(25-x^2))/(15x^2) is 1/15 (-arcsin(1/5 x)-(sqrt(25-x^2))/x)+C