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

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

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integral of x^{-3}cos(1/x)

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

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

Solution steps

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

Simplify

x^{- 3} cos (\frac{1}{x}) : \frac{cos ( \frac{1}{x} )}{x ^{3}}

= \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

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