∫xcos⁡x+12x3esin⁡x+x2dx

I=∫(xcos⁡x+1)esin⁡xdxesin⁡xx2xesin⁡x+1 Put 2xesin⁡x+1=t2⇒ 2esin⁡x+esin⁡xcos⁡xxdx=2tdt⇒ esin⁡x(1+xcos⁡x)dx=tdt∴ I=∫tdtt2−12⋅t I=2∫dtt2−1=log⁡t−1t+1+C,where t2=2xesin⁡x+1