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

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

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

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

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

Solution steps

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

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

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

Apply u-substitution

: \int - sin (u) d u

= \int - sin (u) d u

Take the constant out:

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

= - \int sin (u) d u

Use the common integral:

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

= - (- cos (u))

Substitute back

u = \frac{1}{x}

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

Simplify

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

Add a constant to the solution

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

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