What is the integral of (sin(6/x))/(x^2) ?

The integral of (sin(6/x))/(x^2) is 1/6 cos(6/x)+C

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integral of (sin(6/x))/(x^2)

\int \frac{sin ( \frac{6}{x} )}{x ^{2}} d x

\frac{1}{6} cos (\frac{6}{x}) + C

Solution steps

\int \frac{sin ( \frac{6}{x} )}{x ^{2}} d x

Apply u-substitution

: \int - \frac{sin ( u )}{6} d u

= \int - \frac{sin ( u )}{6} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

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

= - \frac{1}{6} (- cos (u))

Substitute back

u = \frac{6}{x}

= - \frac{1}{6} (- cos (\frac{6}{x}))

Simplify

= \frac{1}{6} cos (\frac{6}{x})

Add a constant to the solution

= \frac{1}{6} cos (\frac{6}{x}) + C

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What is the integral of (sin(6/x))/(x^2) ?
The integral of (sin(6/x))/(x^2) is 1/6 cos(6/x)+C