What is the integral of 1/(x^2sqrt(36-x^2)) ?

The integral of 1/(x^2sqrt(36-x^2)) is -(sqrt(36-x^2))/(36x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(x^2sqrt(36-x^2))

\int \frac{1}{x ^{2} 36 - x ^{2}} d x

- \frac{36 - x ^{2}}{36 x} + C

Solution steps

\int \frac{1}{x ^{2} 36 - x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{cot ( u )}{36 cos ( u ) sin ( u )} d u

= \int \frac{cot ( u )}{36 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}{36} \cdot \int \frac{cot ( u )}{cos ( u ) sin ( u )} d u

Simplify

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

= \frac{1}{36} \cdot \int csc^{2} (u) d u

Use the common integral:

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

= \frac{1}{36} (- cot (u))

Substitute back

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

= \frac{1}{36} (- cot (arcsin (\frac{1}{6} x)))

Simplify

\frac{1}{36} (- cot (arcsin (\frac{1}{6} x))) : - \frac{36 - x ^{2}}{36 x}

= - \frac{36 - x ^{2}}{36 x}

Add a constant to the solution

= - \frac{36 - x ^{2}}{36 x} + C

\int_{- π}^{π} sin (x) cos (x) d x

y^{^{'}} = x^{2} + x, y (1) = 3

\int sin^{3} (6 x) cos^{2} (6 x) d x

\int_{0}^{3} π (\frac{1}{x + 5})^{2} d x

y = 3 x + 1, y = x + 1

What is the integral of 1/(x^2sqrt(36-x^2)) ?
The integral of 1/(x^2sqrt(36-x^2)) is -(sqrt(36-x^2))/(36x)+C