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

The integral of (sec(1/x)tan(1/x))/(x^2) is -sec(1/x)+C

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integral of (sec(1/x)tan(1/x))/(x^2)

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

- sec (\frac{1}{x}) + C

Solution steps

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

Apply u-substitution

: \int - sec (u) tan (u) d u

= \int - sec (u) tan (u) d u

Take the constant out:

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

= - \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= - \int 1 d v

Integral of a constant:

\int a d x = a x

= - 1 \cdot v

Substitute back

= - 1 \cdot sec (\frac{1}{x})

Simplify

= - sec (\frac{1}{x})

Add a constant to the solution

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

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What is the integral of (sec(1/x)tan(1/x))/(x^2) ?
The integral of (sec(1/x)tan(1/x))/(x^2) is -sec(1/x)+C