What is the integral of (x-9)sin(pix) ?

The integral of (x-9)sin(pix) is -1/pi cos(pix)(x-9)+1/(pi^2)sin(pix)+C

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integral of (x-9)sin(pix)

\int (x - 9) sin (π x) d x

- \frac{1}{π} cos (π x) (x - 9) + \frac{1}{π ^{2}} sin (π x) + C

Solution steps

\int (x - 9) sin (π x) d x

Apply Integration By Parts

: - \frac{1}{π} cos (π x) (x - 9) - \int - \frac{1}{π} cos (π x) d x

= - \frac{1}{π} cos (π x) (x - 9) - \int - \frac{1}{π} cos (π x) d x

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

= - \frac{1}{π} cos (π x) (x - 9) - (- \frac{1}{π ^{2}} sin (π x))

Simplify

= - \frac{1}{π} cos (π x) (x - 9) + \frac{1}{π ^{2}} sin (π x)

Add a constant to the solution

= - \frac{1}{π} cos (π x) (x - 9) + \frac{1}{π ^{2}} sin (π x) + C

(\frac{d y}{d x}) + 6 y = e^{- x}

f (x) = 5 sin (x)

\frac{d}{d x} (\frac{x ^{2} + 5 x}{x + 2})

y^{^{''}} - 4 y^{^{'}} + y = 0

\int \frac{1}{9 + x ^{2}} d x

What is the integral of (x-9)sin(pix) ?
The integral of (x-9)sin(pix) is -1/pi cos(pix)(x-9)+1/(pi^2)sin(pix)+C