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

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

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integral of (-sin(1/x))/(x^3)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - u sin (u) d u

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{1}{x}

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

Simplify

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

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

Add a constant to the solution

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

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