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

## Expert Answers 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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