[mathjax]
積分問題32番
問題
解答
\(\displaystyle\int\displaystyle\frac{x}{1+\cos x}dx\)
\(=\displaystyle\int\displaystyle\frac{x(1-\cos x)}{(1+\cos x)(1-\cos x)}dx\)
\(=\displaystyle\int\biggl(\displaystyle\frac{x}{\sin^2 x}\)\(-\displaystyle\frac{x\cos x}{\sin^2 x}\biggr)dx\)
第一項
部分積分する。
\(\displaystyle\int\displaystyle\frac{x}{\sin^2 x}dx\)
\(=-\displaystyle\frac{x}{\tan x}+\displaystyle\int\displaystyle\frac{dx}{\tan x}\)
\(=-\displaystyle\frac{x}{\tan x}+\log|\sin x|+C\)
第二項
こちらも部分積分。
\(-\displaystyle\int\displaystyle\frac{x\cos x}{\sin^2 x} dx\)
\(=\displaystyle\frac{x}{\sin x}-\displaystyle\int\displaystyle\frac{dx}{\sin x}\)
\(=\displaystyle\frac{x}{\sin x}-\log\biggl|\tan\displaystyle\frac{x}{2}\biggr|+C\)
まとめ
\(\displaystyle\int\displaystyle\frac{x}{1+\cos x}dx\)
\(=-\displaystyle\frac{x}{\tan x}+\log|\sin x|+\displaystyle\frac{x}{\sin x}-\log\biggl|\tan\displaystyle\frac{x}{2}\biggr|+C\)
\(=\displaystyle\frac{x}{\sin x}-\displaystyle\frac{x}{\tan x}-\log|1+\cos x|+C\)
