CÂU HỎI: Biết rằng \(\int\left(\cos ^{3} x \cdot \sin 3 x+\sin ^{3} x \cdot \cos 3 x\right) \mathrm{d} x=\frac{a}{b} \cos 4 x+C \text { với } a, b \in \mathbb{Z}, \frac{a}{b}\) là phân số tối giản \((a<0;b>0)\). Tính 2a+b.

\(\begin{aligned}
&\cos ^{3} x \cdot \sin 3 x+\sin ^{3} x \cdot \cos 3 x=\cos ^{3} x \cdot\left(3 \sin x-4 \sin ^{3} x\right)+\sin ^{3} x \cdot\left(4 \cos ^{3} x\right. \\
&=3 \cos ^{3} x \cdot \sin x-4 \sin ^{3} x \cdot \cos ^{3} x+4 \sin ^{3} x \cdot \cos ^{3} x+3 \sin ^{3} x \cdot \cos x \\
&=3 \cos ^{3} x \cdot \sin x+3 \sin ^{3} x \cdot \cos x \\
&=3 \sin x \cdot \cos x \cdot\left(\cos ^{2} x-\sin ^{2} x\right) \\
&=\frac{3}{2} \sin 2 x \cdot\left(\cos ^{2} x-\sin ^{2} x\right) \\
&=\frac{3}{2} \sin 2 x \cdot \cos 2 x=\frac{3}{4} \sin 4 x
\end{aligned}\)

\(\begin{aligned}
&\Rightarrow \int\left(\cos ^{3} x \cdot \sin 3 x+\sin ^{3} x \cdot \cos 3 x\right) \mathrm{d} x=\frac{3}{4} \int \sin 4 x \mathrm{~d} x=-\frac{3}{16} \cos 4 x+C=\frac{a}{b} \cos 4 x+C \\
&\text { Vì: }\left\{\begin{array}{l}
a<0 \\
b>0
\end{array}\right.\Rightarrow\left\{\begin{array}{l}
a=-3 \\
b=16
\end{array} \Rightarrow 2 a+b=10 .\right.
\end{aligned}\)