What is the integral of 1/(t^2sqrt(6-t^2)) ?

The integral of 1/(t^2sqrt(6-t^2)) is -(sqrt(6-t^2))/(6t)+C

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integral of 1/(t^2sqrt(6-t^2))

\int \frac{1}{t ^{2} 6 - t ^{2}} d t

- \frac{6 - t ^{2}}{6 t} + C

Solution steps

\int \frac{1}{t ^{2} 6 - t ^{2}} d t

Apply Trigonometric Substitution

= \int \frac{cot ( u )}{6 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}{6} \cdot \int \frac{cot ( u )}{sin ( u ) cos ( u )} d u

Simplify

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

= \frac{1}{6} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{6} (- cot (u))

Substitute back

u = arcsin (\frac{1}{6} t)

= \frac{1}{6} (- cot (arcsin (\frac{1}{6} t)))

Simplify

\frac{1}{6} (- cot (arcsin (\frac{1}{6} t))) : - \frac{6 - t ^{2}}{6 t}

= - \frac{6 - t ^{2}}{6 t}

Add a constant to the solution

= - \frac{6 - t ^{2}}{6 t} + C

x \to 0 lim (\frac{tan ( 8 x )}{x})

\int \frac{x - 2}{x + 1} d x

\int cos (6 x) sin (6 x) d x

\int_{0}^{1} (\frac{6}{1 + x ^{2}}) d x

\frac{d}{d x} ((4 - x)^{2})

What is the integral of 1/(t^2sqrt(6-t^2)) ?
The integral of 1/(t^2sqrt(6-t^2)) is -(sqrt(6-t^2))/(6t)+C