tan(x/2)=tを用いる三角関数の積分

\(\tan\,\dfrac{x}{2}=t\) とおくと

\({\sin}\,x=\dfrac{2t}{1+t^2}\)

\({\cos}\,x=\dfrac{1-t^2}{1+t^2}\)

\(\dfrac{dt}{dx}=\dfrac{1}{2{\cos}^2\,\dfrac{x}{2}}=\dfrac{1}{2}(1+{\tan}^2\,\dfrac{x}{2})=\dfrac{1}{2}(1+t^2)\)

\(dx=\dfrac{2}{1+t^2}dt\)

以上より,

\(\displaystyle \int\dfrac{1}{1+\sin\,x+\cos\,x}dx\)

\(\displaystyle =\int\dfrac{1}{1+\dfrac{2t}{1+t^2}+\dfrac{1-t^2}{1+t^2}}\cdot\dfrac{2}{1+t^2}dt\)

\(\displaystyle =\int\dfrac{2}{2t+2}dt\)

\(\displaystyle =\int\dfrac{1}{t+1}dt\)

\(=\log |t+1|+C\)

\(=\log |\tan\,\dfrac{x}{2}+1|+C\)　(\(C\)は積分定数)