E =4\Bigl[12\cdot\tfrac{\sin^{4}t}{4} +\bigl(-\cos t+\tfrac{\cos^{3}t}{3}\bigr)\Bigr]_{0}^{\pi} =\bigl[12\sin^{4}t -4\cos t +\tfrac{4}{3}\cos^{3}t\bigr]_{0}^{\pi} =(-4\cdot(-1)+\tfrac{4}{3}(-1)) -( -4\cdot1+\tfrac{4}{3}\,1) =0.
Passo 6
O termo A é \displaystyle A=\int_{0}^{\pi}4\sin^{2}t\cos t\ln(1+4\sin^{2}t)\,dt.
Note que sob t\mapsto\pi-t temos \sin^{2}(\pi-t)=\sin^{2}t, \cos(\pi-t)=-\cos t e \ln(1+4\sin^{2}t) invariante,
logo o integrando muda de sinal e A=0.
Passo 7
Agora
I =2\Bigl[A+B+C-\tfrac12D -E\Bigr] =2\Bigl[0+0+(-\pi)-\tfrac12\cdot\tfrac12\ln5 -0\Bigr] =2\Bigl[-\pi -\tfrac14\ln5\Bigr] =-2\pi -\tfrac12\ln5.
Revisando o cálculo de D, como D=\tfrac12\ln5, o termo em I é -\tfrac12D=-\tfrac14\ln5.
Resposta
\int_{\gamma}\vec F\cdot d\vec r =-2\pi \;-\;\tfrac12\ln5.