Cho hàm số $f(x)$ có $f\left(\dfrac{\pi}{2}\right)=0$ và...

Ta có $f(x)=\int \cos 2 x \cdot \sin ^3 x \mathrm{~d} x=\int\left(2 \cos ^2 x-1\right)\left(1-\cos ^2 x\right) \sin x \mathrm{~d} x$.
Đặt $t=\cos x \Rightarrow \mathrm{d} t=-\sin x \mathrm{~d} x$.
Khi đó: $\int\left(2 \cos ^2 x-1\right)\left(1-\cos ^2 x\right) \sin x \mathrm{~d} x=-\int\left(2 t^2-1\right)\left(1-t^2\right) \mathrm{d} t=\int\left(2 t^4-3 t^2+1\right) \mathrm{d} t$ $=\dfrac{2}{5} t^5-t^3+t+C$
$=\dfrac{2}{5} \cos ^5 x-\cos ^3 x+\cos x+C$
Suy ra: $f(x)=\dfrac{2}{5} \cos ^5 x-\cos ^3 x+\cos x+C$
Mà $f\left(\dfrac{\pi}{2}\right)=0 \Leftrightarrow C=0$.
Do đó $f(x)=\dfrac{2}{5} \cos ^5 x-\cos ^3 x+\cos x=\cos x\left(\dfrac{2}{5} \cos ^4 x-\cos ^2 x+1\right)=\cos x\left[\dfrac{2}{5}\left(1-\sin ^2 x\right)^2+\right.$ $\left.\sin ^2 x\right]$
$I=\int_0^{\dfrac{\pi}{6}} f(x) \mathrm{d} x=\int_0^{\dfrac{\pi}{6}} \cos x\left[\dfrac{2}{5}\left(1-\sin ^2 x\right)^2+\sin ^2 x\right] \mathrm{d} x$
Đặt $t=\sin x \Rightarrow \mathrm{d} t=\cos x \mathrm{~d} x$.
Đổi cận $\left\{\begin{array}{l}x=0 \Rightarrow t=0 \\ x=\dfrac{\pi}{6} \Rightarrow t=\dfrac{1}{2}\end{array}\right.$
$I=\int_0^{\dfrac{1}{2}}\left[\dfrac{2}{5}\left(1-t^2\right)^2+t^2\right] \mathrm{d} t=\dfrac{1}{5} \int_0^{\dfrac{1}{2}}\left(2 t^4+t^2+2\right) \mathrm{d} t=\left.\dfrac{1}{5}\left(\dfrac{2}{5} t^5+\dfrac{1}{3} t^3+2 t\right)\right|_0 ^{\dfrac{1}{2}}=\dfrac{253}{1200}$