Wednesday, December 4, 2013

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

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`


...

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.`

No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Exposition A "decidedly amused" Bobby Gillian leaves the offices of Tolman & Sharp where he is given an envelope containing $1...