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

The integral of (tan(2/x))/(x^2) is 1/2 ln|cos(2/x)|+C

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

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

\frac{1}{2} ln cos (\frac{2}{x}) + C

Solution steps

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

Apply u-substitution

: \int - \frac{tan ( u )}{2} d u

= \int - \frac{tan ( 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} \cdot \int tan (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1}{v} d v

= - \frac{1}{2} \cdot \int - \frac{1}{v} d v

Take the constant out:

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

= - \frac{1}{2} (- \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{2} (- ln ∣ v ∣)

Substitute back

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

Simplify

= \frac{1}{2} ln cos (\frac{2}{x})

Add a constant to the solution

= \frac{1}{2} ln cos (\frac{2}{x}) + C

n = 1 \sum \infty \frac{2 n + 1}{7 ^{n}}

\int_{- 1}^{1} tan (x) d x

x \to 1 - lim (- x^{2} + 5 x + a)

\int_{- π}^{π} sin^{2} (x) d x

(3, 5), (1, - 7)

What is the integral of (tan(2/x))/(x^2) ?
The integral of (tan(2/x))/(x^2) is 1/2 ln|cos(2/x)|+C