a) Ta có: \({\sin ^4}x = \dfrac{{{{\left( {1 - \cos 2x} \right)}^2}}}{4}\)\( = \dfrac{1}{4}\left( {1 - 2\cos 2x + {{\cos }^2}2x} \right)\)
\( = \dfrac{1}{4}\left( {1 - 2\cos 2x + \dfrac{{1 + \cos 4x}}{2}} \right)\) \( = \dfrac{1}{4}\left( {\dfrac{3}{2} - 2\cos 2x + \dfrac{1}{2}\cos 4x} \right)\)
Khi đó \(\int {{{\sin }^4}x} dx\)\( = \int {\dfrac{1}{4}\left( {\dfrac{3}{2} - 2\cos 2x + \dfrac{1}{2}\cos 4x} \right)dx} \) \( = \int {\left( {\dfrac{3}{8} - \dfrac{1}{2}\cos 2x + \dfrac{1}{8}\cos 4x} \right)dx} \)
\( = \dfrac{3}{8}x - \dfrac{1}{2}.\dfrac{{\sin 2x}}{2} + \dfrac{1}{8}.\dfrac{{\sin 4x}}{4} + C\) \( = \dfrac{3}{8}x - \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}} + C\)
b) Ta có: \(\int {\dfrac{1}{{{{\sin }^3}x}}dx} \)\( = \int {\dfrac{{\sin x}}{{{{\sin }^4}x}}dx} \) \( = \int {\dfrac{{\sin x}}{{{{\left( {1 - {{\cos }^2}x} \right)}^2}}}dx} \)
Đặt \(t = \cos x \Rightarrow dt = - \sin xdx\) ta có:
c) \(\int {{{\sin }^3}x{{\cos }^4}xdx} \)
Đặt \(t = \cos x \Rightarrow dt = - \sin xdx\).
Khi đó \(\int {{{\sin }^3}x{{\cos }^4}xdx} \)\( = \int {{{\sin }^2}x.{{\cos }^4}x.\sin xdx} \) \( = \int {\left( {1 - {t^2}} \right).{t^4}.\left( { - dt} \right)} \)
\( = \int {\left( { - {t^4} + {t^6}} \right)dt} \) \( = - \dfrac{{{t^5}}}{5} + \dfrac{{{t^7}}}{7} + C\) \( = - \dfrac{{{{\cos }^5}x}}{5} + \dfrac{{{{\cos }^7}x}}{7} + C\).
d) Ta có: \({\sin ^4}x{\cos ^4}x\)\( = {\left( {\dfrac{1}{2}\sin 2x} \right)^4} = \dfrac{1}{{{2^4}}}{\sin ^4}2x\) \( = \dfrac{1}{{16}}{\left( {{{\sin }^2}2x} \right)^2} = \dfrac{1}{{16}}.{\left[ {\dfrac{{1 - \cos 4x}}{2}} \right]^2}\)
\( = \dfrac{1}{{64}}{\left( {1 - \cos 4x} \right)^2}\) \( = \dfrac{1}{{64}}\left( {1 - 2\cos 4x + {{\cos }^2}4x} \right)\) \( = \dfrac{1}{{64}} - \dfrac{1}{{32}}\cos 4x + \dfrac{1}{{64}}.\dfrac{{1 + \cos 8x}}{2}\)
\( = \dfrac{3}{{128}} - \dfrac{1}{{32}}\cos 4x + \dfrac{1}{{128}}\cos 8x\)
