`int (x - 1) sin(pi x ) dx` Evaluate the integral

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You need to solve the integral `int (x-1) sin (pi*x) dx = int x*sin (pi*x) dx - int sin (pi*x)dx`

You need to use substitution `pi*x = t => pi*dx = dt => dx = (dt)/(pi)`

`int x*sin (pi*x) dx = 1/(pi^2) int t*sin t`

You need to use the integration by parts for `int t*sin t `  such that:

`int udv = uv - int vdu`

`u = t => du = dt`

`dv = sin t=> v = -cos t`

` `

`int t*sin t = -t*cos t + int cos t dt`

`1/(pi^2) int t*sin t = 1/(pi^2)(-t*cos t + sin t) + c`

Replacing back the variable yields:

`int x*sin (pi*x) dx = 1/(pi^2)(-pi*x*cos(pi*x) + sin (pi*x)) + c`

`int (x-1) sin (pi*x) dx = 1/(pi^2)(-pi*x*cos(pi*x) + sin (pi*x))+ (cos (pi*x))/(pi) + c`

Hence, evaluating the integral, using  integration by parts, yields `int (x-1) sin (pi*x) dx = 1/(pi^2)(-pi*x*cos(pi*x) + sin (pi*x))+ (cos (pi*x))/(pi) + c.`

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