What is the integral of (1-x)sin(npix) ?

The integral of (1-x)sin(npix) is -1/(pin)cos(pinx)+1/(pin)xcos(pinx)-1/(pi^2n^2)sin(pinx)+C

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integral of (1-x)sin(npix)

\int (1 - x) sin (nπ x) d x

- \frac{1}{πn} cos (πn x) + \frac{1}{πn} x cos (πn x) - \frac{1}{π ^{2} n ^{2}} sin (πn x) + C

Solution steps

\int (1 - x) sin (nπ x) d x

Expand

(1 - x) sin (nπ x) : sin (πn x) - x sin (πn x)

= \int sin (πn x) - x sin (πn x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int sin (πn x) d x - \int x sin (πn x) d x

\int sin (πn x) d x = - \frac{1}{πn} cos (πn x)

\int x sin (πn x) d x = - \frac{1}{πn} x cos (πn x) + \frac{1}{π ^{2} n ^{2}} sin (πn x)

= - \frac{1}{πn} cos (πn x) - (- \frac{1}{πn} x cos (πn x) + \frac{1}{π ^{2} n ^{2}} sin (πn x))

Simplify

= - \frac{1}{πn} cos (πn x) + \frac{1}{πn} x cos (πn x) - \frac{1}{π ^{2} n ^{2}} sin (πn x)

Add a constant to the solution

= - \frac{1}{πn} cos (πn x) + \frac{1}{πn} x cos (πn x) - \frac{1}{π ^{2} n ^{2}} sin (πn x) + C

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What is the integral of (1-x)sin(npix) ?
The integral of (1-x)sin(npix) is -1/(pin)cos(pinx)+1/(pin)xcos(pinx)-1/(pi^2n^2)sin(pinx)+C