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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 12: 21) tính nguyên hàm $\int$cos4x.sinx dx 22) tính nguyên hàm $\int$cos3x.cos2x dx 23) tính nguyên hàm $\int$sin5x.sin2x dx

Toán Lớp 12: 21) tính nguyên hàm
$\int$cos4x.sinx dx
22) tính nguyên hàm
$\int$cos3x.cos2x dx
23) tính nguyên hàm
$\int$sin5x.sin2x dx

Comments ( 2 )

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

  2. • 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. $\displaystyle\int \sin 5x. \sin 2x dx$
    $= \displaystyle\int \dfrac{-1}{2} (\cos 7x – \cos 3x) dx$
    $= \dfrac{-1}{2}. \Big( \dfrac{\sin 7x}{7} – \dfrac{\sin 3x}{3} \Big) + C$
    $= \dfrac{-\sin 7x}{14} + \dfrac{\sin 3x}{6} + C$

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222-9+11+12:2*14+14 = ? ( )