What is the integral of-(2ycos(pi/y))/pi ?

The integral of-(2ycos(pi/y))/pi is -1/pi y^2cos(pi/y)+ysin(pi/y)-pi\Ci(pi/y)+C

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integral of-(2ycos(pi/y))/pi

\int - \frac{2 y cos ( \frac{π}{y} )}{π} d y

- \frac{1}{π} y^{2} cos (\frac{π}{y}) + y sin (\frac{π}{y}) - π Ci (\frac{π}{y}) + C

Solution steps

\int - \frac{2 y cos ( \frac{π}{y} )}{π} d y

Take the constant out:

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

= - \frac{2}{π} \cdot \int y cos (\frac{π}{y}) d y

Apply Integration By Parts

: \frac{1}{2} y^{2} cos (\frac{π}{y}) - \int \frac{π}{2} sin (\frac{π}{y}) d y

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

\int \frac{π}{2} sin (\frac{π}{y}) d y = - \frac{π ^{2}}{2} (- \frac{1}{π} y sin (\frac{π}{y}) + Ci (\frac{π}{y}))

= - \frac{2}{π} (\frac{1}{2} y^{2} cos (\frac{π}{y}) - (- \frac{π ^{2}}{2} (- \frac{1}{π} y sin (\frac{π}{y}) + Ci (\frac{π}{y}))))

Simplify

- \frac{2}{π} (\frac{1}{2} y^{2} cos (\frac{π}{y}) - (- \frac{π ^{2}}{2} (- \frac{1}{π} y sin (\frac{π}{y}) + Ci (\frac{π}{y})))) : - \frac{1}{π} y^{2} cos (\frac{π}{y}) + y sin (\frac{π}{y}) - π Ci (\frac{π}{y})

= - \frac{1}{π} y^{2} cos (\frac{π}{y}) + y sin (\frac{π}{y}) - π Ci (\frac{π}{y})

Add a constant to the solution

= - \frac{1}{π} y^{2} cos (\frac{π}{y}) + y sin (\frac{π}{y}) - π Ci (\frac{π}{y}) + C

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What is the integral of-(2ycos(pi/y))/pi ?
The integral of-(2ycos(pi/y))/pi is -1/pi y^2cos(pi/y)+ysin(pi/y)-pi\Ci(pi/y)+C