1.

inte^x(cotx+Insinx)dx

Answer»

Solution :`inte^x(COTX+Insinx)DX`
[Integrating by parts TAKING `e^x` as FIRST function and cotx as SECOND function.]
=`e^x.Insinx-inte^x.Insinxdx+inte^xInsinxdx+C`
=`e^xInsinx+C`


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