What is the integral of (cos(1/x))/(x^3) ?

The integral of (cos(1/x))/(x^3) is -(sin(1/x))/x-cos(1/x)+C

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integral of (cos(1/x))/(x^3)

\int \frac{cos ( \frac{1}{x} )}{x ^{3}} d x

- \frac{sin ( \frac{1}{x} )}{x} - cos (\frac{1}{x}) + C

Solution steps

\int \frac{cos ( \frac{1}{x} )}{x ^{3}} d x

Apply u-substitution

: \int - u cos (u) d u

= \int - u cos (u) d u

Take the constant out:

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

= - \int u cos (u) d u

Apply Integration By Parts

: u sin (u) - \int sin (u) d u

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

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

= - (u sin (u) - (- cos (u)))

Substitute back

u = \frac{1}{x}

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

Simplify

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

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

Add a constant to the solution

= - \frac{sin ( \frac{1}{x} )}{x} - cos (\frac{1}{x}) + C

y^{^{'}} = 3 x - y

y^{^{'}} = \frac{x ^{2} y ^{4} - x ^{2}}{y ^{3}}

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

\frac{( 35 + 16 )}{s ^{2} - s - 6}

\int \frac{5}{x ^{2} + 9} d x

What is the integral of (cos(1/x))/(x^3) ?
The integral of (cos(1/x))/(x^3) is -(sin(1/x))/x-cos(1/x)+C