$\sin^4 x +\cos^4 x =(\sin^2 x +\cos^2 x)^2 -2\sin^2 x \cos^2 x =1-\dfrac{1}{2}\sin^2 2x =1-\dfrac{1}{4}(1-\cos 4x)$
$=\dfrac{3}{4}+\dfrac{1}{4}\cos 4x$
$I=\dfrac{3}{4} \int e^{2x} dx+\dfrac{1}{4}\int e^{2x} .\cos 4x dx=\dfrac{3}{8}e^{2x} +\dfrac{1}{4}I_1$
Tính $I_1$ đặt $\cos 4x = u \Rightarrow \dfrac{1}{4}\sin 4x dx = du$ và $e^{2x}dx=dv \Rightarrow \dfrac{1}{2}e^{2x}=v$
$I_1= \dfrac{1}{2}e^{2x}\cos 4x -\dfrac{1}{8}\int e^{2x} \sin 4x dx$
Tính $I_2= \int e^{2x}\sin 4x dx$ đặt $\sin 4x = u \Rightarrow -\dfrac{1}{4}\cos 4x dx=du$ và $e^{2x}dx=dv \Rightarrow \dfrac{1}{2}e^{2x}=v$
$I_2=\dfrac{1}{2}e^{2x}\sin 4x +\dfrac{1}{4}\int e^{2x}\cos 4x dx=\dfrac{1}{2}e^{2x}\sin 4x +\dfrac{1}{4}I_1$
$I_1= \dfrac{1}{2}e^{2x}\cos 4x -\dfrac{1}{8} \bigg (\dfrac{1}{2}e^{2x}\sin 4x +\dfrac{1}{4}I_1 \bigg)$
tự thay vào làm nốt nhé