Đáp án:

\(\begin{array}{l}
21)\quad \displaystyle\int\cos4x.\sin xdx
= \dfrac16\cos3x - \dfrac{1}{10}\cos5x + C\\
22)\quad \displaystyle\int\cos3x\cos2xdx
= \dfrac{1}{10}\sin5x + \dfrac12\sin x + C\\
23)\quad \displaystyle\int\sin5x.\sin2xdx
= \dfrac{1}{6}\sin3x - \dfrac{1}{14}\sin7x + C
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}\text{(Áp dụng công thức biến đổi tích thành tổng)}\\
21)\quad \displaystyle\int\cos4x.\sin xdx\\
= \dfrac12\displaystyle\int(\sin5x - \sin3x)dx\\
= \dfrac12\left(-\dfrac15\cos5x + \dfrac13\cos3x + C\right)\\
= \dfrac16\cos3x - \dfrac{1}{10}\cos5x + C\\
22)\quad \displaystyle\int\cos3x\cos2xdx\\
= \dfrac12\displaystyle\int(\cos5x + \cos x)dx\\
= \dfrac12\left(\dfrac15\sin5x + \sin x + C\right)dx\\
= \dfrac{1}{10}\sin5x + \dfrac12\sin x + C\\
23)\quad \displaystyle\int\sin5x.\sin2xdx\\
= \dfrac12\displaystyle\int(\cos3x - \cos7x)dx\\
= \dfrac12\left(\dfrac13\sin3x - \dfrac17\sin7x + C\right)\\
= \dfrac{1}{6}\sin3x - \dfrac{1}{14}\sin7x + C
\end{array}\) 

• Họ nguyên hàm : 

21. $\displaystyle\int \cos 4x. \sin x dx$

$= \displaystyle\int \dfrac{1}{2} (\sin 5x - \sin 3x) dx$

$= \dfrac{1}{2} \Big( \displaystyle\int \sin 5x - \displaystyle\int \sin 3x \Big) dx$

$= \dfrac{1}{2}.\Big( \dfrac{- \cos 5x}{5} + \dfrac{ \cos 3x}{3} \Big) + C$

$= \dfrac{- \cos 5x}{10} + \dfrac{ \cos 3x}{6} +C$

22. $\displaystyle\int \cos 3x. \cos 2x dx$

$= \displaystyle\int \dfrac{1}{2} (\cos 5x + \cos x) dx$

$= \dfrac{1}{2} \Big( \dfrac{ \sin 5x}{5} + \dfrac{\sin x} {1} \Big) + C$

$= \dfrac{\sin 5x}{10} + \dfrac{\sin x}{2} + C$

23. $\displaystyle\int \sin 5x. \sin 2x dx$

$= \displaystyle\int \dfrac{-1}{2} (\cos 7x - \cos 3x) dx$

$= \dfrac{-1}{2}. \Big( \dfrac{\sin 7x}{7} - \dfrac{\sin 3x}{3} \Big) + C$

$= \dfrac{-\sin 7x}{14} + \dfrac{\sin 3x}{6} + C$