Tuesday, December 21, 2010

`int sin(pi t) dt` Evaluate the indefinite integral.

You need to use the following substitution `pi*t = u` , such that:


`pi*t = u=> pi*dt= du => dt= (du)/(pi)`


`int sin(pi*t)dt= (1/(pi))*int sin u du`


`(1/(pi))*int sin u du= - cos u + c`


Replacing back  `pi*t` for u yields:


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


Hence, evaluating the indefinite integral, yields `int sin(pi*t)dt = -(cos(pi*t))/(pi) + c`

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