What is the integral of 1/(t^2sqrt(4-t^2)) ?

The integral of 1/(t^2sqrt(4-t^2)) is -(sqrt(4-t^2))/(4t)+C

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integral of 1/(t^2sqrt(4-t^2))

\int \frac{1}{t ^{2} 4 - t ^{2}} d t

- \frac{4 - t ^{2}}{4 t} + C

Solution steps

\int \frac{1}{t ^{2} 4 - t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{4 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{4 cos ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{4} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{4} (- cot (u))

Substitute back

u = arcsin (\frac{1}{2} t)

= \frac{1}{4} (- cot (arcsin (\frac{1}{2} t)))

Simplify

\frac{1}{4} (- cot (arcsin (\frac{1}{2} t))) : - \frac{4 - t ^{2}}{4 t}

= - \frac{4 - t ^{2}}{4 t}

Add a constant to the solution

= - \frac{4 - t ^{2}}{4 t} + C

\frac{\partial}{\partial x} (- 13 e^{x} sin (y) - 14 y)

\frac{d}{d a} (sin (a) - sin (6 a) + sin (5 a))

x \to 0 lim (e^{- 4 x})

y^{^{'}} = \frac{1.5 - 0.8 y}{0.5 + 0.8 x}

\int x (e^{x^{2}} + 2) d x

What is the integral of 1/(t^2sqrt(4-t^2)) ?
The integral of 1/(t^2sqrt(4-t^2)) is -(sqrt(4-t^2))/(4t)+C