最近更新于 2025-08-19 00:05

形式一

\begin{aligned}
\int\csc\ x\ dx&=\int\frac1{\sin x}dx\\
&\xlongequal{二倍角公式}\int\frac1{2\sin\frac{x}2\cos\frac{x}2}dx\\
&\xlongequal{u=\frac{x}2}\int\frac1{\sin u\cos u} du\\
&\xlongequal{\sin u=\tan u\cos u}\int\frac1{\cos^2u\tan u}du\\
&\xlongequal{d(\tan u)=\sec^2u=\frac1{\cos^2u}}\int\frac1{\tan u}d\tan u\\
&=\ln|\tan u|+C\\
&=\ln|\tan\frac{x}2|+C
\end{aligned}

形式二

裂项参考：https://blog.iyatt.com/?p=20837

\begin{aligned}
\int\csc\ x\ dx&=\int\frac1{\sin x}dx\\
&=\int\frac{\sin x}{\sin^2x}dx\\
&\xlongequal{d\cos x=-\sin x}-\int\frac1{\sin^2x}d\cos x\\
&\xlongequal{\sin^2x+\cos^2x=1}\int\frac1{\cos^2x-1}d\cos x\\
&=\int\frac1{(\cos x+1)(\cos x-1)}d\cos x\\
&\xlongequal{裂项}\frac12\int(\frac1{\cos x-1}-\frac1{\cos x+1})d\cos x\\
&=\frac12(\ln|\cos x - 1|)-\ln(\cos x+1)+C\\
&=\frac12\ln|\frac{\cos x-1}{\cos x+1}|+C

\end{aligned}

a