What is the integral of (cos(x^{-1}))/(x^2) ?

The integral of (cos(x^{-1}))/(x^2) is -sin(1/x)+C

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integral of (cos(x^{-1}))/(x^2)

\int \frac{cos ( x ^{- 1} )}{x ^{2}} d x

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

Solution steps

\int \frac{cos ( x ^{- 1} )}{x ^{2}} d x

x^{- 1} = \frac{1}{x}

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

Apply u-substitution

: \int - cos (u) d u

= \int - cos (u) d u

Take the constant out:

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

= - \int cos (u) d u

Use the common integral:

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

= - sin (u)

Substitute back

u = \frac{1}{x}

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

Add a constant to the solution

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

\int \frac{x ^{2} + 2}{3} d x

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\int x^{2} sin (y x) d x

\int (x + cos (x))^{'} d x

\int x sech^{2} (x) d x

What is the integral of (cos(x^{-1}))/(x^2) ?
The integral of (cos(x^{-1}))/(x^2) is -sin(1/x)+C