$I=\int \limits_0^{\frac{\pi}{4}} \dfrac{5\cos x -4\sin x}{\cos^3 x (\tan x +1)^3}dx +\int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{5\cos x -4\sin x}{\sin^3 x (\cot x +1)^3}dx$
$=\int \dfrac{5-4\tan x}{(\tan x +1)^3}d(\tan x) -\int \dfrac{5\cot x -4}{(\cot x +1)^3}d(\cot x)$
$=\int \dfrac{5-4t}{(t+1)^3}dt - \int \dfrac{5u-4}{(u+1)^3}du$
$=\int \bigg ( \dfrac{9}{(t+1)^3}- \dfrac{4}{(t+1)^2}\bigg) dt - \int \bigg (\dfrac{5}{(u+1)^2}-\dfrac{9}{(u+1)^3} \bigg ) du$ Dễ rồi nhé