[mathjax]
積分問題86番
問題
解答
\(\displaystyle\int\displaystyle\frac{x}{\sin^2 x}dx\)
\(=-\displaystyle\frac{x}{\tan x}+\displaystyle\int\displaystyle\frac{dx}{\tan x}\) \(cdots\) 部分積分
\(=-\displaystyle\frac{x}{\tan x}+\displaystyle\int\displaystyle\frac{\cos x}{\sin x}dx\)
\(=-\displaystyle\frac{x}{\tan x}+\log|\sin x|+C\)
