What is the integral of 1/(y^2sqrt(5-y^2)) ?

The integral of 1/(y^2sqrt(5-y^2)) is -(sqrt(5-y^2))/(5y)+C

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integral of 1/(y^2sqrt(5-y^2))

\int \frac{1}{y ^{2} 5 - y ^{2}} d y

- \frac{5 - y ^{2}}{5 y} + C

Solution steps

\int \frac{1}{y ^{2} 5 - y ^{2}} d y

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{5 sin ( u ) cos ( u )} d u

= \int \frac{cot ( u )}{5 sin ( u ) cos ( u )} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{cot ( u )}{sin ( u ) cos ( u )} d u

Simplify

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

= \frac{1}{5} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{5} (- cot (u))

Substitute back

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

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

Simplify

\frac{1}{5} (- cot (arcsin (\frac{1}{5} y))) : - \frac{5 - y ^{2}}{5 y}

= - \frac{5 - y ^{2}}{5 y}

Add a constant to the solution

= - \frac{5 - y ^{2}}{5 y} + C

\frac{d}{d x} (\frac{x sin ( x )}{1 + cos ( x )})

\int_{1}^{e} \frac{4}{z} d z

\frac{\partial}{\partial y} (x^{\frac{1}{y}})

\frac{d ^{2}}{d x ^{2}} y + 64 y = 0

\frac{\partial}{\partial x} (\frac{2 y}{x ^{2} - y ^{2}})

What is the integral of 1/(y^2sqrt(5-y^2)) ?
The integral of 1/(y^2sqrt(5-y^2)) is -(sqrt(5-y^2))/(5y)+C