What is the integral of 1/(t^2sqrt(289-t^2)) ?

The integral of 1/(t^2sqrt(289-t^2)) is -(sqrt(289-t^2))/(289t)+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(289-t^2))

\int \frac{1}{t ^{2} 289 - t ^{2}} d t

- \frac{289 - t ^{2}}{289 t} + C

Solution steps

\int \frac{1}{t ^{2} 289 - t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{289 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{289 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}{289} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{289} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{289} (- cot (u))

Substitute back

u = arcsin (\frac{1}{17} t)

= \frac{1}{289} (- cot (arcsin (\frac{1}{17} t)))

Simplify

\frac{1}{289} (- cot (arcsin (\frac{1}{17} t))) : - \frac{289 - t ^{2}}{289 t}

= - \frac{289 - t ^{2}}{289 t}

Add a constant to the solution

= - \frac{289 - t ^{2}}{289 t} + C

\int_{- 7}^{7} \frac{1}{2} (49 - x^{2}) d x

t \to π lim (sin (\frac{t}{2}))

\frac{10}{s \cdot ( s + 1 ) \cdot ( s + 10 )}

2 t

What is the integral of 1/(t^2sqrt(289-t^2)) ?
The integral of 1/(t^2sqrt(289-t^2)) is -(sqrt(289-t^2))/(289t)+C