What is the integral of 1/(t^2sqrt(64-t^2)) ?

The integral of 1/(t^2sqrt(64-t^2)) is -(sqrt(64-t^2))/(64t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(t^2sqrt(64-t^2))

\int \frac{1}{t ^{2} 64 - t ^{2}} d t

- \frac{64 - t ^{2}}{64 t} + C

Solution steps

\int \frac{1}{t ^{2} 64 - t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{64 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{64 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}{64} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{64} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{64} (- cot (u))

Substitute back

u = arcsin (\frac{1}{8} t)

= \frac{1}{64} (- cot (arcsin (\frac{1}{8} t)))

Simplify

\frac{1}{64} (- cot (arcsin (\frac{1}{8} t))) : - \frac{64 - t ^{2}}{64 t}

= - \frac{64 - t ^{2}}{64 t}

Add a constant to the solution

= - \frac{64 - t ^{2}}{64 t} + C

3 y^{^{''}} + 4 y^{^{'}} = 0

\int_{1}^{e} x^{4} ln (x) d x

\int_{0}^{π} 4 sin (x) d x

\frac{\partial}{\partial y} (2 x ln (y))

x \to 0 + lim (2 - \frac{3}{x ^{\frac{1}{3}}})

What is the integral of 1/(t^2sqrt(64-t^2)) ?
The integral of 1/(t^2sqrt(64-t^2)) is -(sqrt(64-t^2))/(64t)+C