How to integrate (sinx)(cosx)(e^cosx)dx ?

Question

3687 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-15T19:56:29

#-cosxe^cosx-e^cosx+C#

#cosx=z#
#-sinxdx=dz#

#intsinxcosxe^cosx=int-ze^zdz#

#intudv=uv-intvdu#

#-z=u##color(blue)rarr##-dz=du#

#e^zdz=dv##color(blue)rarr##e^z=v#

#int-ze^zdz#=#-ze^z-int-e^zdz#

=#-ze^z-e^z+C#

Reverse the substitution

#intsinxcosxe^cosx==-cosxe^cosx-e^cosx+C#

Answer 2 · 2018-04-15T20:03:40

#= -\ e^( cos x) (cos x - 1 ) + C #

#int (sinx)(cosx)(e^cosx)dx#

By IBP:

#= - int d (e^cosx) \ cosx #

#= - (cos x \ e^( cos x) - int e^cosx d(cosx) ) #

Remember that: #int e^Q \ d Q= e^Q + C#, so we have:

#= - (cos x \ e^( cos x) - e^cosx + C ) #

#= -\ e^( cos x) (cos x - 1 ) + C #