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inte^x(cotx+Insinx)dx |
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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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