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2015-04-09T03:13:19+05:30

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We are given:

I= -\int\limits^a_b {x^2\ 2Cos\ 2x} \, dx \\\\let\ u=x^2,\ \ v=2Cos\ 2x\\ \int\limits^{}_{} {v} \, dx =\int\limits^{}_{} {2Cos\ 2x} \, dx =Sin\ 2x\\\\Integration\ by\ parts\\\\\int\limits^{}_{} {u\ v} \, dx =u\ \int\limits^{}_{} {v} \, dx\ -\  \int\limits^{}_{} {u'\ (\int\limits^{}_{} {v} \, dx)} \, dx \\\\=>I=-x^2\ Sin\ 2x\ +\  \int\limits^{}_{} {(2x)\ Sin2x} \, dx\\=>\ \ =-x^2\ Sin2x\ +\ (x)\ (-Cos2x)- \int\limits^{}_{} {(1)*(-Cos2x)} \, dx \\\\=>\ \ =\ -x^2\ Sin2x-x\ Cos2x+ \frac{1}{2} Sin2x\ +K

we applied the integration by parts rule , the second time for integration  of 
           2 x * sin 2x  by  taking   u = x  and  v = 2 Sin2x.

 finally, the integration involved just  Cos 2x.

So we are able reduce the power of x  1 each time we applied the rule.

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