Câu 3 dài quá tôi làm tắt k gõ cận nhé
Đặt $x=-t$
$I =\int \dfrac{2008^t (\sin^6 t +\cos^6 t)}{2008^t+1}dt =\int \dfrac{2008^x (\sin^6 x +\cos^6 x)}{2008^x+1}dx$
$\Rightarrow 2I =\int \dfrac{2008^x (\sin^6 x +\cos^6 x)}{2008^x+1}dx + \int \dfrac{\sin^6 x +\cos^6 x}{2008^x+1}dx$
$=\int (\sin^6 x +\cos^6 x)dx$
Ta có $\sin^6 x + \cos^6 x= (\sin^2 x + \cos^2 x)^3 -3\sin^2 x \cos^2 x ( \sin^2 x +\cos^2 x)=1-3\sin^2 x \cos^2 x$
$= 1-\dfrac{3}{4}\sin^2 2x=\dfrac{5}{8} +\dfrac{3}{8}\cos 4x$
Vậy $2I = \int (\dfrac{5}{8} +\dfrac{3}{8}\cos 4x)dx $ dễ rồi nhé