Question

Calcule a seguinte integral.

Answer 1

           $\displaystyle I=\int \dfrac{1}{1-\cos x+\sin x}dx$

Hagamos el siguiente cambio de variable

          $u=\tan \dfrac{x}{2}\to x=2\arctan(u) \to dx =\dfrac{2du}{1+u^2}\\ \\ \cos x=\dfrac{1-u^2}{1+u^2}\;,\; \sin x =\dfrac{2u}{1+u^2}$

Sustituimos en la integral

         $\displaystyle I=\int \dfrac{1}{1-\dfrac{1-u^2}{1+u^2}+\dfrac{2u}{1+u^2}}\cdot \dfrac{2}{1+u^2}du\\ \\ \\ I=\int \dfrac{2}{(1+u^2)-(1-u^2)+2u}\,du\\ \\ \\ I=\int \dfrac{du}{u^2+u}\\ \\ I=\int \dfrac{du}{u(u+1)}\\ \\ I=\int \dfrac{1}{u}-\dfrac{1}{u+1}du\\ \\ \\ \boxed{I=\ln \left|\dfrac{u}{u+1}\right|}$


Por ello

        $\displaystyle \boxed{\boxed{\int \dfrac{1}{1-\cos x+\sin x}dx=\ln \left|\dfrac{\tan(x/2)}{\tan(x/2)+1}\right|+C}}$

Answer 2

Resolver a integral indefinida

$\displaystyle\int{\dfrac{1}{1-\cos(x)+\mathrm{sen\,}(x)}\,dx}$


A substituição adequada é

$\bullet\;\;u=\mathrm{tg\,}(\frac{x}{2})\;\;\Rightarrow\;\;x=2\,\mathrm{arc\,tg\,}(u)\;\;\Rightarrow\;\;dx=\dfrac{2}{u^{2}+1}\,du\;\;\;\;\;\mathbf{(i)}\\ \\ \\ \bullet\;\;u=\dfrac{\mathrm{sen\,}(\frac{x}{2})}{\cos (\frac{x}{2})}$


Como $\cos (\frac{x}{2})\neq 0,$ podemos multiplicar o numerador e o denominador por $2\cos (\frac{x}{2}):$

$u=\dfrac{2\,\mathrm{sen\,}(\frac{x}{2})\cos(\frac{x}{2})}{2\cos^{2} (\frac{x}{2})}\\ \\ \\ u=\dfrac{\mathrm{sen\,}(x)}{2}\cdot \sec^{2}(\frac{x}{2})\\ \\ \\ u=\dfrac{\mathrm{sen\,}(x)}{2}\cdot \left(\mathrm{tg^{2}\,}(\frac{x}{2})+1 \right )\\ \\ \\ u=\dfrac{\mathrm{sen\,}(x)}{2}\cdot (u^{2}+1)\\ \\ \mathrm{sen\,}(x)=\dfrac{2u}{u^{2}+1}\;\;\;\;\;\mathbf{(ii)}$


$\bullet\;\;\cos (x)=2\cos^{2}(\frac{x}{2})-1\\ \\ \cos (x)=\dfrac{2}{\sec^{2}(\frac{x}{2})}-1\\ \\ \\ \cos (x)=\dfrac{2}{\mathrm{tg^{2}\,}(\frac{x}{2})+1}-1\\ \\ \\ \cos (x)=\dfrac{2}{u^{2}+1}-1\\ \\ \\ \cos (x)=\dfrac{2-(u^{2}+1)}{u^{2}+1}\\ \\ \\ \cos (x)=\dfrac{2-u^{2}-1}{u^{2}+1}\\ \\ \\ \cos (x)=\dfrac{1-u^{2}}{u^{2}+1}\;\;\;\;\;\;\mathbf{(iii)}$


$\bullet\;\;u=\dfrac{2\,\mathrm{sen\,}(\frac{x}{2})\cos(\frac{x}{2})}{2\cos^{2} (\frac{x}{2})}\\ \\ \\ u=\dfrac{2\,\mathrm{sen\,}(\frac{x}{2})\cos(\frac{x}{2})}{(2\cos^{2} (\frac{x}{2})-1)+1}\\ \\ \\ u=\dfrac{\mathrm{sen\,}(x)}{\cos(x)+1}\;\;\;\;\;\;\mathbf{(iv)}$


Efetuando as substituições das equações $\mathbf{(i)},\,\mathbf{(ii)}$ e $\mathbf{(iii)},$ na integral, obtemos

$\displaystyle\int{\dfrac{1}{1-\frac{1-u^{2}}{u^{2}+1}+\frac{2u}{u^{2}+1}}\cdot \dfrac{2}{u^{2}+1}\,du}\\ \\ \\ =\displaystyle\int{\dfrac{2}{(u^{2}+1)-(1-u^{2})+(2u)}\,du}\\ \\ \\ =\displaystyle\int{\dfrac{2}{u^{2}+1-1+u^{2}+2u}\,du}\\ \\ \\ =\displaystyle\int{\dfrac{2}{2u^{2}+2u}\,du}\\ \\ \\ =\displaystyle\int{\dfrac{\diagup\!\!\!\! 2}{\diagup\!\!\!\! 2u\,(u+1)}\,du}\\ \\ \\ =\displaystyle\int{\dfrac{1}{u\,(u+1)}\,du}$


Decompondo a função integranda em frações parciais:

$\dfrac{1}{u\,(u+1)}=\dfrac{A}{u}+\dfrac{B}{u+1}\\ \\ \\ \dfrac{1}{u\,(u+1)}=\dfrac{A(u+1)+Bu}{u(u+1)}\\ \\ \\ \dfrac{1}{u\,(u+1)}=\dfrac{(A+B)u+A}{u(u+1)}\\ \\ \\ 1=(A+B)u+A$


Para satisfazer a igualdade entre os polinômios, devemos ter

$A=1\,\text{ e }\,B=-1.$


Então, a integral fica

$\displaystyle\int{\dfrac{1}{u\,(u+1)}\,du}\\ \\ \\ =\displaystyle\int{\left(\dfrac{1}{u}-\dfrac{1}{u+1} \right )\,du}\\ \\ \\ =\mathrm{\ell n\,}|u|-\mathrm{\ell n\,}|u+1|+C$


Substituindo de volta para a variável $x,$ utilizando a equação $\mathbf{(iv)},$ temos

$=\mathrm{\ell n\,}\left|\dfrac{\mathrm{sen\,}(x)}{\cos(x)+1}\right|-\mathrm{\ell n\,}\left|\dfrac{\mathrm{sen\,}(x)}{\cos(x)+1}+1\right|+C\\ \\ \\ =\mathrm{\ell n\,}\left|\dfrac{\mathrm{sen\,}(x)}{\cos(x)+1}\right|-\mathrm{\ell n\,}\left|\dfrac{\mathrm{sen\,}(x)+\cos(x)+1}{\cos(x)+1}\right|+C$


Simplificando os logaritmos, finalmente chegamos a

$\bullet\;\;\displaystyle\int{\dfrac{1}{1-\cos(x)+\mathrm{sen}(x)}\,dx}=\mathrm{\ell n\,}\left|\dfrac{\mathrm{sen\,}(x)}{\mathrm{sen\,}(x)+\cos(x)+1}\right|+C$