Решение интегралов 11кл Полное решение = лучший

Решите задачу:

$\int \frac{6sin^2x}{4+3cos2x} dx=\int \frac{6\cdot \frac{1-cos2x}{2}}{4+3\cdot cos2x} dx=\int \frac{3(1-cos2x)}{4+3cos2x} dx=\\\\=[t=tgx,\; cos2x=\frac{1-t^2}{1+t^2} \; ,\; dx=\frac{dt}{1+t^2}\; ]=\\\\=\int \frac{3(1-\frac{1-t^2}{1+t^2})}{4+3\cdot \frac{1-t^2}{1+t^2}} \cdot \frac{dt}{1+t^2}=\int \frac{3(1+t^2-1+t^2)}{4+4t^2+3-3t^2}\cdot \frac{dt}{1+t^2} =\int \frac{6t^2}{(7+t^2)(1+t^2)} =I\\\\\\\frac{6t^2}{(7+t^2)(1+t^2)}=\frac{At+B}{7+t^2}+\frac{Ct+D}{1+t^2} \\\\6t^2=(At+B)(1+t^2)+(Ct+D)(7+t^2)$

$6t^2=At+Bt^2+At^3+B+7Ct+7D+Ct^3+Dt^2\\\\t^3\; |\; A+C=0\; ,\; \; A=-C\\\\t^2\; |\; B+D=6\\\\t\; |\; \; A+7C=0\; ,\; -C+7C=0\; ,\; 6C=0\; ,\; C=0\; \to \; A=0\\\\t^0\; |\; B+7D=0\\\\(B+D)-(B+7D)=6-0\; \; \to \; \; -6D=6\; ,\; \; D=-1\\\\B=6-D=6+1=7\\\\I=\int \frac{7}{7+t^2} dt+\int \frac{-1}{1+t^2} dt=7\cdot \frac{1}{\sqrt7}\cdot arctg(\frac{t}{\sqrt7})-arctgt+C=\\\\=\sqrt7\cdot arctg(\frac{tgx}{\sqrt7})-arctg(tgx)+C=\sqrt7\cdot arctg(\frac{tgx}{\sqrt7})-x+C$