Thứ 1: Ta có $\cos 4x-\cos 2x = -2\sin^2 x (2\cos 2x +1)$
Do đó $\dfrac{x (\cos 4x-\cos 2x )+\sin 2x}{\sin^4 x (2\cos 2x+1)}=\dfrac{-2x \sin^2 x (2\cos 2x+1)}{\sin^4 x(2\cos 2x +1)}+\dfrac{\sin 2x}{\sin^4 x (2\cos 2x +1)}$
$=-\dfrac{2x}{\sin^2 x}+\dfrac{2\cos x}{\sin^3 x \bigg [ 2(1-2\sin^2 x) +1) \bigg ] }=-\dfrac{2x}{\sin^2 x}+\dfrac{2\cos x}{\sin^3 x \bigg [ 3-4\sin^2 x \bigg ] }$
Vậy $I = -2\int \dfrac{x}{\sin^2 x}dx+ \int\dfrac{2\cos x}{\sin^3 x \bigg [ 3-4\sin^2 x \bigg ] }dx$
$=2I_0+2\int \dfrac{d(\sin x)}{\sin^3 x (3-4\sin^2 x)}$
Trong đó $I_1=\int \dfrac{d(\sin x)}{\sin^3 x (3-4\sin^2 x)}=\int \dfrac{1}{t^3(3-4t^2)}dt$
$=\int \bigg (\dfrac{1}{3t^3}+\dfrac{4}{9t}-\dfrac{16t}{9(4t^2 -3)} \bigg ) dt$
Đều là những tp khá cơ bản
Trong đó $\dfrac{16t}{9(4t^2 -3)} =\dfrac{4}{9(\sqrt 3 +2t)}-\dfrac{4}{9(\sqrt 3 -2t)}$
$I_0 = -\int \dfrac{x}{\sin^2 x}dx$ đặt $x=u \Rightarrow dx = du$ và $-\dfrac{1}{\sin^2 x}dx = dv \Rightarrow \cot x = v$
$\Rightarrow I_0 = x\cot x -\int \cot x dx$ e tự tính nốt nha