Friday, April 9, 2010

`int x sin(x^2) dx` Evaluate the indefinite integral.

You need to evaluate the indefinite integral by performing the substitution `x^2 = t` , such that:


`x^2 = t => 2xdx = dt => xdx = (dt)/2`


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


`(1/2)int sin t dt= (1/2)(-cos t) + c`


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Replacing back `x^2` for t yields:


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


Hence, evaluating the indefinite integral yields `int x*sin(x^2) dx=  -(1/2)(cos x^2) + c`

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