Integral (4x -3) sin (2x +5)dx Semoga ada yg bisa membantu trimkasih

Integral (4x -3) sin (2x +5)dx

u= (4x -3)

du=4dx

dv=sin (2x +5)dx

v=integral sin (2x +5)dx

=-½cos(2x +5)

Integral (4x -3) sin (2x +5)dx

=-½Integral udv

=-½(uv-integral vdu)

=(4x -3)(-½cos(2x +5))–integral(-½cos(2x +5) ) 4dx

=-½(4x -3)cos(2x +5)+2integral(cos(2x +5) ) dx

=-½(4x -3)cos(2x +5)+sin(2x +5) +c

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