What is the integral of 8/(x^2)cos(8/x) ?

The integral of 8/(x^2)cos(8/x) is -sin(8/x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 8/(x^2)cos(8/x)

\int \frac{8}{x ^{2}} cos (\frac{8}{x}) d x

- sin (\frac{8}{x}) + C

Solution steps

\int \frac{8}{x ^{2}} cos (\frac{8}{x}) d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{2}} cos (\frac{8}{x}) d x

Apply u-substitution

: \int - \frac{cos ( u )}{8} d u

= 8 \cdot \int - \frac{cos ( u )}{8} d u

Take the constant out:

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

= 8 (- \frac{1}{8} \cdot \int cos (u) d u)

Use the common integral:

\int cos (u) d u = sin (u)

= 8 (- \frac{1}{8} sin (u))

Substitute back

u = \frac{8}{x}

= 8 (- \frac{1}{8} sin (\frac{8}{x}))

Simplify

8 (- \frac{1}{8} sin (\frac{8}{x})) : - sin (\frac{8}{x})

= - sin (\frac{8}{x})

Add a constant to the solution

= - sin (\frac{8}{x}) + C

\frac{\partial}{\partial x} (y cos (x + y))

\int 15 x \cdot 1 - x d x

\int \frac{cos ( 4 x ) cos ( 4 x )}{4} d x

\frac{4 - sec ( x )}{tan ( x )}

\frac{d}{d s} (\frac{s}{s ^{2} + 2})

What is the integral of 8/(x^2)cos(8/x) ?
The integral of 8/(x^2)cos(8/x) is -sin(8/x)+C