What is the integral of cos(1/x)cos(3/x) 1/(x^2) ?

The integral of cos(1/x)cos(3/x) 1/(x^2) is 1/2 (-1/4 sin(4/x)-1/2 sin(2/x))+C

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integral of cos(1/x)cos(3/x) 1/(x^2)

\int cos (\frac{1}{x}) cos (\frac{3}{x}) \frac{1}{x ^{2}} d x

\frac{1}{2} (- \frac{1}{4} sin (\frac{4}{x}) - \frac{1}{2} sin (\frac{2}{x})) + C

Solution steps

\int cos (\frac{1}{x}) cos (\frac{3}{x}) \frac{1}{x ^{2}} d x

Rewrite using trig identities

= \int \frac{cos ( \frac{4}{x} ) + cos ( - \frac{2}{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{cos ( \frac{4}{x} ) + cos ( - \frac{2}{x} )}{x ^{2}} d x

Expand

\frac{cos ( \frac{4}{x} ) + cos ( - \frac{2}{x} )}{x ^{2}} : \frac{cos ( \frac{4}{x} )}{x ^{2}} + \frac{cos ( - \frac{2}{x} )}{x ^{2}}

= \frac{1}{2} \cdot \int \frac{cos ( \frac{4}{x} )}{x ^{2}} + \frac{cos ( - \frac{2}{x} )}{x ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{cos ( \frac{4}{x} )}{x ^{2}} d x + \int \frac{cos ( - \frac{2}{x} )}{x ^{2}} d x)

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

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

= \frac{1}{2} (- \frac{1}{4} sin (\frac{4}{x}) - \frac{1}{2} sin (\frac{2}{x}))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{4} sin (\frac{4}{x}) - \frac{1}{2} sin (\frac{2}{x})) + C

What is the integral of cos(1/x)cos(3/x) 1/(x^2) ?
The integral of cos(1/x)cos(3/x) 1/(x^2) is 1/2 (-1/4 sin(4/x)-1/2 sin(2/x))+C