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

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

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integral of 1/(x^2sin(pi/x))

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

- \frac{1}{π} ln tan (\frac{π}{2 x}) + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{π sin ( u )} d u

= \int - \frac{1}{π sin ( u )} d u

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= - \frac{1}{π} ln ∣ v ∣

Substitute back

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

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

Add a constant to the solution

= - \frac{1}{π} ln tan (\frac{π}{2 x}) + C

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