Sustitución de Weierstrass

Por las identidades y fórmulas de trigonometría,

${\displaystyle {\begin{aligned}\operatorname {sen} x&=2\operatorname {sen} \left({\frac {x}{2}}\right)\cos \left({\frac {x}{2}}\right)\\&=2\left({\frac {t}{\sqrt {t^{2}+1}}}\right)\left({\frac {1}{\sqrt {t^{2}+1}}}\right)\\&={\frac {2t}{t^{2}+1}}\end{aligned}}}$

y

${\displaystyle {\begin{aligned}\cos x&=2\cos ^{2}\left({\frac {x}{2}}\right)-1\\&={\frac {2}{t^{2}+1}}-1\\&={\frac {2-(t^{2}+1)}{t^{2}+1}}\\&={\frac {1-t^{2}}{1+t^{2}}}\end{aligned}}}$

Finalmente, si ${\displaystyle t=\tan \left({\frac {x}{2}}\right)}$ entonces

${\displaystyle {\begin{aligned}dt&={\frac {1}{2}}\sec ^{2}\left({\frac {x}{2}}\right)dx\\&={\frac {dx}{2\cos ^{2}\left({\frac {x}{2}}\right)}}\\&={\frac {dx}{2\left({\frac {1}{t^{2}+1}}\right)}}\end{aligned}}}$

Por lo tanto

${\displaystyle dx={\frac {2}{t^{2}+1}}dt}$

${\displaystyle {\begin{aligned}\int \csc x\,dx&=\int {\frac {dx}{\operatorname {sen} x}}\\[6pt]&=\int \left({\frac {1+t^{2}}{2t}}\right)\left({\frac {2}{1+t^{2}}}\right)dt&&t=\tan {\frac {x}{2}}\\[6pt]&=\int {\frac {dt}{t}}\\[6pt]&=\ln |t|+C\\[6pt]&=\ln \left|\tan {\frac {x}{2}}\right|+C.\end{aligned}}}$

Se puede confirmar el resultado anterior usando un método estándar para evaluar la integral de la cosecante multiplicando el numerador y el denominador por ${\displaystyle \csc x-\cot x}$ y realizando el siguiente cambio de variable

${\displaystyle {\begin{aligned}u&=\csc x-\cot x\\du&=(-\csc x\cot x+\csc ^{2}x)dx\end{aligned}}}$

Por lo que

${\displaystyle {\begin{aligned}\int \csc x\,dx&=\int {\frac {\csc x(\csc x-\cot x)}{\csc x-\cot x}}\,dx\\[6pt]&=\int {\frac {(\csc ^{2}x-\csc x\cot x)}{\csc x-\cot x}}\;dx\\[6pt]&=\int {\frac {du}{u}}\\[6pt]&=\ln |u|+C\\[6pt]&=\ln |\csc x-\cot x|+C\end{aligned}}}$

Ahora, las fórmulas del ángulo mitad para senos y cosenos son

${\displaystyle \operatorname {sen} ^{2}\theta ={\frac {1-\cos 2\theta }{2}}\quad {\text{y}}\quad \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}}$

respectivamente y permiten obtener

${\displaystyle {\begin{aligned}\int \csc x\,dx&=\ln \left|\tan {\frac {x}{2}}\right|+C=\ln {\sqrt {\frac {1-\cos x}{1+\cos x}}}+C\\[6pt]&=\ln {\sqrt {{\frac {1-\cos x}{1+\cos x}}\cdot {\frac {1-\cos x}{1-\cos x}}}}+C\\[6pt]&=\ln {\sqrt {\frac {(1-\cos x)^{2}}{\operatorname {sen} ^{2}x}}}+C\\[6pt]&=\ln {\sqrt {\left({\frac {1-\cos x}{\operatorname {sen} x}}\right)^{2}}}+C\\[6pt]&=\ln {\sqrt {\left({\frac {1}{\operatorname {sen} x}}-{\frac {\cos x}{\operatorname {sen} x}}\right)^{2}}}+C\\[6pt]&=\ln {\sqrt {(\csc x-\cot x)^{2}}}+C\\[6pt]&=\ln \left|\csc x-\cot x\right|+C\end{aligned}}}$

por lo que los dos resultados son equivalentes. La expresión

${\displaystyle \tan \left({\frac {x}{2}}\right)={\frac {1-\cos x}{\operatorname {sen} x}}}$

es una de las fórmulas de la tangente del ángulo mitad. La integral de la secante puede evaluarse de manera similar.

${\displaystyle {\begin{aligned}\int _{0}^{2\pi }{\frac {dx}{2+\cos x}}&=\int _{0}^{\pi }{\frac {dx}{2+\cos x}}+\int _{\pi }^{2\pi }{\frac {dx}{2+\cos x}}\\[6pt]&=\int _{0}^{\infty }{\frac {2\,dt}{3+t^{2}}}+\int _{-\infty }^{0}{\frac {2\,dt}{3+t^{2}}}&t&=\tan {\frac {x}{2}}\\[6pt]&=\int _{-\infty }^{\infty }{\frac {2\,dt}{3+t^{2}}}\\[6pt]&={\frac {2}{\sqrt {3}}}\int _{-\infty }^{\infty }{\frac {du}{1+u^{2}}}&t&=u{\sqrt {3}}\\[6pt]&={\frac {2\pi }{\sqrt {3}}}.\end{aligned}}}$

En la primera línea, no se sustituye simplemente ${\displaystyle t=0}$ por ambos límites de integración. Se debe tener en cuenta la singularidad (en este caso, una asíntota vertical) de ${\displaystyle t=\tan {\frac {x}{2}}}$ en ${\displaystyle x=\pi }$. Alternativamente, primero se debe evaluar la integral indefinida y luego aplicar los valores del intervalo.

${\displaystyle {\begin{aligned}\int {\frac {dx}{2+\cos x}}&=\int {\frac {1}{2+{\frac {1-t^{2}}{1+t^{2}}}}}{\frac {2\,dt}{t^{2}+1}}&&t=\tan {\frac {x}{2}}\\[6pt]&=\int {\frac {2\,dt}{2(t^{2}+1)+(1-t^{2})}}\\[6pt]&=\int {\frac {2\,dt}{t^{2}+3}}\\[6pt]&={\frac {2}{3}}\int {\frac {dt}{\left({\frac {t}{\sqrt {3}}}\right)^{2}+1}}&&u={\frac {t}{\sqrt {3}}}\\[6pt]&={\frac {2}{\sqrt {3}}}\int {\frac {du}{u^{2}+1}}&&\tan \theta =u\\[6pt]&={\frac {2}{\sqrt {3}}}\int \cos ^{2}\theta \sec ^{2}\theta \,d\theta \\[6pt]&={\frac {2}{\sqrt {3}}}\int d\theta \\[6pt]&={\frac {2}{\sqrt {3}}}\theta +C\\[6pt]&={\frac {2}{\sqrt {3}}}\arctan \left({\frac {t}{\sqrt {3}}}\right)+C\\[6pt]&={\frac {2}{\sqrt {3}}}\arctan \left[{\frac {\tan(x/2)}{\sqrt {3}}}\right]+C\end{aligned}}}$

Por simetría,

${\displaystyle {\begin{aligned}\int _{0}^{2\pi }{\frac {dx}{2+\cos x}}&=2\int _{0}^{\pi }{\frac {dx}{2+\cos x}}\\&=\lim _{b\rightarrow \pi }{\frac {4}{\sqrt {3}}}\arctan \left({\frac {\tan x/2}{\sqrt {3}}}\right){\Biggl |}_{0}^{b}\\[6pt]&={\frac {4}{\sqrt {3}}}{\Biggl [}\lim _{b\rightarrow \pi }\arctan \left({\frac {\tan b/2}{\sqrt {3}}}\right)-\arctan(0){\Biggl ]}\\&={\frac {4}{\sqrt {3}}}\left({\frac {\pi }{2}}-0\right)={\frac {2\pi }{\sqrt {3}}}\end{aligned}}}$

que es igual al resultado anterior.

${\displaystyle {\begin{aligned}\int {\frac {dx}{a\cos x+b\sin x+c}}&=\int {\frac {2dt}{a(1-t^{2})+2bt+c(t^{2}+1)}}\\[6pt]&=\int {\frac {2dt}{(c-a)t^{2}+2bt+a+c}}\\[6pt]&={\frac {2}{\sqrt {c^{2}-(a^{2}+b^{2})}}}\arctan {\frac {(c-a)\tan {\frac {x}{2}}+b}{\sqrt {c^{2}-(a^{2}+b^{2})}}}+C\end{aligned}}}$

Si ${\displaystyle 4E=4(c-a)(c+a)-(2b)^{2}=4(c^{2}-(a^{2}+b^{2}))>0.}$