What is the integral of 1/(x^2)sin(4/x)cos(4/x) ?

The integral of 1/(x^2)sin(4/x)cos(4/x) is 1/16 cos(8/x)+C

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integral of 1/(x^2)sin(4/x)cos(4/x)

\int \frac{1}{x ^{2}} sin (\frac{4}{x}) cos (\frac{4}{x}) d x

\frac{1}{16} cos (\frac{8}{x}) + C

Solution steps

\int \frac{1}{x ^{2}} sin (\frac{4}{x}) cos (\frac{4}{x}) d x

Rewrite using trig identities

= \int \frac{sin ( \frac{8}{x} )}{2 x ^{2}} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{sin ( \frac{8}{x} )}{x ^{2}} d x

Apply u-substitution

: \int - \frac{sin ( u )}{8} d u

= \frac{1}{2} \cdot \int - \frac{sin ( u )}{8} d u

Take the constant out:

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

= \frac{1}{2} (- \frac{1}{8} \cdot \int sin (u) d u)

Use the common integral:

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

= \frac{1}{2} (- \frac{1}{8} (- cos (u)))

Substitute back

u = \frac{8}{x}

= \frac{1}{2} (- \frac{1}{8} (- cos (\frac{8}{x})))

Simplify

\frac{1}{2} (- \frac{1}{8} (- cos (\frac{8}{x}))) : \frac{1}{16} cos (\frac{8}{x})

= \frac{1}{16} cos (\frac{8}{x})

Add a constant to the solution

= \frac{1}{16} cos (\frac{8}{x}) + C

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What is the integral of 1/(x^2)sin(4/x)cos(4/x) ?
The integral of 1/(x^2)sin(4/x)cos(4/x) is 1/16 cos(8/x)+C