[mathjax]
問題
計算
\(\displaystyle\int\displaystyle\frac{\sin^2 x}{\sin^2 x+1}dx\)
\(=\displaystyle\int\biggl(1-\displaystyle\frac{1}{\sin^2 x+1}\biggr)dx\)
\(=x-\displaystyle\int\displaystyle\frac{\tan^2 x+1}{2\tan^2 x+1}dx\)
※第二項の変形
\(\displaystyle\frac{1}{\sin^2 x+1}\)
\(=\displaystyle\frac{1}{\sin^2 x\biggl(1+\displaystyle\frac{1}{\sin^2 x}\biggr)}\)
\(=\displaystyle\frac{1}{\sin^2 x}\times \displaystyle\frac{1}{1+\displaystyle\frac{1}{\sin^2 x}}\)
\(=\biggl(1+\displaystyle\frac{1}{\tan^2 x}\biggr)\times \displaystyle\frac{1}{1+\biggl(1+\displaystyle\frac{1}{\tan^2 x}\biggr)}\)
\(=\displaystyle\frac{\tan^2 x+1}{2\tan^2 x+1}\)
\(=x-\displaystyle\int\displaystyle\frac{dt}{2t^2+1}\) ※\(t=\tan x\)
\(=x-\displaystyle\frac{1}{2}\displaystyle\int\displaystyle\frac{dt}{t^2+\biggl(\displaystyle\frac{1}{\sqrt{2}}\biggr)^2}\)
\(=x-\displaystyle\frac{\sqrt{2}}{2}\tan^{-1} \sqrt{2}t+C\)
第二項は以下を利用した。
\(=x-\displaystyle\frac{\sqrt{2}}{2}\tan^{-1} (\sqrt{2}\tan x)+C\)
答え
\(x-\displaystyle\frac{\sqrt{2}}{2}\tan^{-1} (\sqrt{2}\tan x)+C\)
