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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1 + cos ( u )}{2} d u

= \frac{1}{2} \cdot \int - \frac{1 + cos ( u )}{2} 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}{2} \cdot \int 1 + cos (u) d u)

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} (- \frac{1}{2} (\int 1 d u + \int cos (u) d u))

\int 1 d u = u

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

= \frac{1}{2} (- \frac{1}{2} (u + sin (u)))

Substitute back

u = \frac{2}{x}

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

Simplify

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

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

Add a constant to the solution

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

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