问题及解答

求

\[\int\frac{1}{\sin^2 x\cos x}dx.\]

[Hint]

\[
\int\frac{1}{\sin^2 x\cos x}dx=\int\sec x\csc^2 xdx=-\int\sec x d\cot x
\]

然后用分部积分.

\[
\begin{split}
\int\frac{1}{\sin^2 x\cos x}dx&=\int\sec x\csc^2 xdx=-\int\sec x d\cot x\\
&=-\sec x\cot x+\int\cot x\sec x\tan x dx\\
&=-\sec x\cot x+\int\sec xdx
\end{split}
\]

回忆

\[
\int\sec xdx=\int\frac{1}{\cos x}dx=\int\frac{\cos x}{\cos^2 x}dx=\int\frac{d\sin x}{1-\sin^2 x}=...
\]