Toán Mọi người giúp mình với \(\int {\dfrac{1}{{{{\sin }^3}x}}dx} \) 13/09/2021 By Charlie Mọi người giúp mình với \(\int {\dfrac{1}{{{{\sin }^3}x}}dx} \)
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: \(\int {\dfrac{{\sin x}}{{{{\left( {1 – {{\cos }^2}x} \right)}^2}}}dx} \)\( = \int {\dfrac{{ – dt}}{{{{\left( {1 – {t^2}} \right)}^2}}}} \) \( = – \int {{{\left[ {\dfrac{1}{{\left( {1 – t} \right)\left( {1 + t} \right)}}} \right]}^2}dt} \) \( = – \dfrac{1}{4}\int {{{\left( {\dfrac{1}{{1 – t}} + \dfrac{1}{{1 + t}}} \right)}^2}dt} \) \( = – \dfrac{1}{4}\int {\left[ {\dfrac{1}{{{{\left( {1 – t} \right)}^2}}} + \dfrac{2}{{\left( {1 – t} \right)\left( {1 + t} \right)}} + \dfrac{1}{{{{\left( {1 + t} \right)}^2}}}} \right]dt} \) \( = – \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\left( {1 – t} \right)}^2}}}} – \dfrac{1}{4}\int {\left( {\dfrac{1}{{1 – t}} + \dfrac{1}{{1 + t}}} \right)dt} – \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\left( {1 + t} \right)}^2}}}} \) \( = – \dfrac{1}{4}.\dfrac{1}{{1 – t}} + \dfrac{1}{4}\ln \left| {1 – t} \right| – \dfrac{1}{4}\ln \left| {1 + t} \right| + \dfrac{1}{4}.\dfrac{1}{{1 + t}} + C\) \( = \dfrac{1}{4}\left( {\dfrac{1}{{1 + t}} – \dfrac{1}{{1 – t}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{1 – t}}{{1 + t}}} \right| + C\) \( = \dfrac{1}{4}\left( {\dfrac{1}{{1 + \cos x}} – \dfrac{1}{{1 – \cos x}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{1 – \cos x}}{{1 + \cos x}}} \right| + C\) \( = \dfrac{1}{4}.\dfrac{{ – 2\cos x}}{{1 – {{\cos }^2}x}} + \dfrac{1}{4}\ln \left| {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} \right| + C\) \( = – \dfrac{{\cos x}}{{2{{\sin }^2}x}} + \dfrac{1}{4}\ln \left( {{{\tan }^2}\dfrac{x}{2}} \right) + C\) \( = – \dfrac{{\cos x}}{{2{{\sin }^2}x}} + \dfrac{1}{2}\ln \left| {\tan \dfrac{x}{2}} \right| + C\). Trả lời
\(\int {\dfrac{1}{{{{\sin }^3}x}}dx} \)\( = \int {\dfrac{{\sin x}}{{{{\sin }^4}x}}dx} \) \( = \int {\dfrac{{\sin x}}{{{{\left( {1 – {{\cos }^2}x} \right)}^2}}}dx} \) Đặt \(t = \cos x \Rightarrow dt = – \sin xdx\) ta có: \(\int {\dfrac{{\sin x}}{{{{\left( {1 – {{\cos }^2}x} \right)}^2}}}dx} \)\( = \int {\dfrac{{ – dt}}{{{{\left( {1 – {t^2}} \right)}^2}}}} \) \( = – \int {{{\left[ {\dfrac{1}{{\left( {1 – t} \right)\left( {1 + t} \right)}}} \right]}^2}dt} \) \( = – \dfrac{1}{4}\int {{{\left( {\dfrac{1}{{1 – t}} + \dfrac{1}{{1 + t}}} \right)}^2}dt} \) \( = – \dfrac{1}{4}\int {\left[ {\dfrac{1}{{{{\left( {1 – t} \right)}^2}}} + \dfrac{2}{{\left( {1 – t} \right)\left( {1 + t} \right)}} + \dfrac{1}{{{{\left( {1 + t} \right)}^2}}}} \right]dt} \) \( = – \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\left( {1 – t} \right)}^2}}}} – \dfrac{1}{4}\int {\left( {\dfrac{1}{{1 – t}} + \dfrac{1}{{1 + t}}} \right)dt} – \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\left( {1 + t} \right)}^2}}}} \) \( = – \dfrac{1}{4}.\dfrac{1}{{1 – t}} + \dfrac{1}{4}\ln \left| {1 – t} \right| – \dfrac{1}{4}\ln \left| {1 + t} \right| + \dfrac{1}{4}.\dfrac{1}{{1 + t}} + C\) \( = \dfrac{1}{4}\left( {\dfrac{1}{{1 + t}} – \dfrac{1}{{1 – t}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{1 – t}}{{1 + t}}} \right| + C\) \( = \dfrac{1}{4}\left( {\dfrac{1}{{1 + \cos x}} – \dfrac{1}{{1 – \cos x}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{1 – \cos x}}{{1 + \cos x}}} \right| + C\) Trả lời
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:
\(\int {\dfrac{{\sin x}}{{{{\left( {1 – {{\cos }^2}x} \right)}^2}}}dx} \)\( = \int {\dfrac{{ – dt}}{{{{\left( {1 – {t^2}} \right)}^2}}}} \) \( = – \int {{{\left[ {\dfrac{1}{{\left( {1 – t} \right)\left( {1 + t} \right)}}} \right]}^2}dt} \) \( = – \dfrac{1}{4}\int {{{\left( {\dfrac{1}{{1 – t}} + \dfrac{1}{{1 + t}}} \right)}^2}dt} \)
\( = – \dfrac{1}{4}\int {\left[ {\dfrac{1}{{{{\left( {1 – t} \right)}^2}}} + \dfrac{2}{{\left( {1 – t} \right)\left( {1 + t} \right)}} + \dfrac{1}{{{{\left( {1 + t} \right)}^2}}}} \right]dt} \) \( = – \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\left( {1 – t} \right)}^2}}}} – \dfrac{1}{4}\int {\left( {\dfrac{1}{{1 – t}} + \dfrac{1}{{1 + t}}} \right)dt} – \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\left( {1 + t} \right)}^2}}}} \)
\( = – \dfrac{1}{4}.\dfrac{1}{{1 – t}} + \dfrac{1}{4}\ln \left| {1 – t} \right| – \dfrac{1}{4}\ln \left| {1 + t} \right| + \dfrac{1}{4}.\dfrac{1}{{1 + t}} + C\) \( = \dfrac{1}{4}\left( {\dfrac{1}{{1 + t}} – \dfrac{1}{{1 – t}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{1 – t}}{{1 + t}}} \right| + C\)
\( = \dfrac{1}{4}\left( {\dfrac{1}{{1 + \cos x}} – \dfrac{1}{{1 – \cos x}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{1 – \cos x}}{{1 + \cos x}}} \right| + C\)
\( = \dfrac{1}{4}.\dfrac{{ – 2\cos x}}{{1 – {{\cos }^2}x}} + \dfrac{1}{4}\ln \left| {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} \right| + C\) \( = – \dfrac{{\cos x}}{{2{{\sin }^2}x}} + \dfrac{1}{4}\ln \left( {{{\tan }^2}\dfrac{x}{2}} \right) + C\)
\( = – \dfrac{{\cos x}}{{2{{\sin }^2}x}} + \dfrac{1}{2}\ln \left| {\tan \dfrac{x}{2}} \right| + C\).
\(\int {\dfrac{1}{{{{\sin }^3}x}}dx} \)\( = \int {\dfrac{{\sin x}}{{{{\sin }^4}x}}dx} \) \( = \int {\dfrac{{\sin x}}{{{{\left( {1 – {{\cos }^2}x} \right)}^2}}}dx} \)
Đặt \(t = \cos x \Rightarrow dt = – \sin xdx\) ta có:
\(\int {\dfrac{{\sin x}}{{{{\left( {1 – {{\cos }^2}x} \right)}^2}}}dx} \)\( = \int {\dfrac{{ – dt}}{{{{\left( {1 – {t^2}} \right)}^2}}}} \) \( = – \int {{{\left[ {\dfrac{1}{{\left( {1 – t} \right)\left( {1 + t} \right)}}} \right]}^2}dt} \) \( = – \dfrac{1}{4}\int {{{\left( {\dfrac{1}{{1 – t}} + \dfrac{1}{{1 + t}}} \right)}^2}dt} \)
\( = – \dfrac{1}{4}\int {\left[ {\dfrac{1}{{{{\left( {1 – t} \right)}^2}}} + \dfrac{2}{{\left( {1 – t} \right)\left( {1 + t} \right)}} + \dfrac{1}{{{{\left( {1 + t} \right)}^2}}}} \right]dt} \) \( = – \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\left( {1 – t} \right)}^2}}}} – \dfrac{1}{4}\int {\left( {\dfrac{1}{{1 – t}} + \dfrac{1}{{1 + t}}} \right)dt} – \dfrac{1}{4}\int {\dfrac{{dt}}{{{{\left( {1 + t} \right)}^2}}}} \)
\( = – \dfrac{1}{4}.\dfrac{1}{{1 – t}} + \dfrac{1}{4}\ln \left| {1 – t} \right| – \dfrac{1}{4}\ln \left| {1 + t} \right| + \dfrac{1}{4}.\dfrac{1}{{1 + t}} + C\) \( = \dfrac{1}{4}\left( {\dfrac{1}{{1 + t}} – \dfrac{1}{{1 – t}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{1 – t}}{{1 + t}}} \right| + C\)
\( = \dfrac{1}{4}\left( {\dfrac{1}{{1 + \cos x}} – \dfrac{1}{{1 – \cos x}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{1 – \cos x}}{{1 + \cos x}}} \right| + C\)