What is the integral of 1/(y^2sqrt(14-y^2)) ?

The integral of 1/(y^2sqrt(14-y^2)) is -(sqrt(14-y^2))/(14y)+C

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integral of 1/(y^2sqrt(14-y^2))

\int \frac{1}{y ^{2} 14 - y ^{2}} d y

- \frac{14 - y ^{2}}{14 y} + C

Solution steps

\int \frac{1}{y ^{2} 14 - y ^{2}} d y

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{14 sin ( u ) cos ( u )} d u

= \int \frac{cot ( u )}{14 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}{14} \cdot \int \frac{cot ( u )}{sin ( u ) cos ( u )} d u

Simplify

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

= \frac{1}{14} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{14} (- cot (u))

Substitute back

u = arcsin (\frac{1}{14} y)

= \frac{1}{14} (- cot (arcsin (\frac{1}{14} y)))

Simplify

\frac{1}{14} (- cot (arcsin (\frac{1}{14} y))) : - \frac{14 - y ^{2}}{14 y}

= - \frac{14 - y ^{2}}{14 y}

Add a constant to the solution

= - \frac{14 - y ^{2}}{14 y} + C

\frac{2}{x}

6 \frac{d y}{d x} + 42 y = 7

\int_{1}^{4} (x + \frac{1}{x}) d x

y = x^{2} - \frac{1}{8} cos (x)

(- 1, 4), (11, - 5)

What is the integral of 1/(y^2sqrt(14-y^2)) ?
The integral of 1/(y^2sqrt(14-y^2)) is -(sqrt(14-y^2))/(14y)+C