What is the integral of xsin((npix)/(40)) ?

The integral of xsin((npix)/(40)) is 1/(pi^2n^2)(-40pinxcos((pinx)/(40))+1600sin((pinx)/(40)))+C

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integral of xsin((npix)/(40))

\int x sin (\frac{nπ x}{40}) d x

\frac{1}{π ^{2} n ^{2}} (- 40 πn x cos (\frac{πn x}{40}) + 1600 sin (\frac{πn x}{40})) + C

Solution steps

\int x sin (\frac{nπ x}{40}) d x

Apply u-substitution

: \int \frac{u sin ( \frac{u}{40} )}{π ^{2} n ^{2}} d u

= \int \frac{u sin ( \frac{u}{40} )}{π ^{2} n ^{2}} d u

Take the constant out:

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

= \frac{1}{π ^{2} n ^{2}} \cdot \int u sin (\frac{u}{40}) d u

Apply Integration By Parts

: - 40 u cos (\frac{u}{40}) - \int - 40 cos (\frac{u}{40}) d u

= \frac{1}{π ^{2} n ^{2}} (- 40 u cos (\frac{u}{40}) - \int - 40 cos (\frac{u}{40}) d u)

\int - 40 cos (\frac{u}{40}) d u = - 1600 sin (\frac{u}{40})

= \frac{1}{π ^{2} n ^{2}} (- 40 u cos (\frac{u}{40}) - (- 1600 sin (\frac{u}{40})))

Substitute back

u = nπ x

= \frac{1}{π ^{2} n ^{2}} (- 40 nπ x cos (\frac{nπ x}{40}) - (- 1600 sin (\frac{nπ x}{40})))

Simplify

= \frac{1}{π ^{2} n ^{2}} (- 40 πn x cos (\frac{πn x}{40}) + 1600 sin (\frac{πn x}{40}))

Add a constant to the solution

= \frac{1}{π ^{2} n ^{2}} (- 40 πn x cos (\frac{πn x}{40}) + 1600 sin (\frac{πn x}{40})) + C

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What is the integral of xsin((npix)/(40)) ?
The integral of xsin((npix)/(40)) is 1/(pi^2n^2)(-40pinxcos((pinx)/(40))+1600sin((pinx)/(40)))+C