Question

Queria saber como resolve: integral dx/senx + tgx
Utilizando o método da substituição de z=tgx/2

Answer 1

$\displaystyle \int\frac{dx}{\sin x+\tan x}\longrightarrow z=\tan \left(\frac{x}{2}\right)\longrightarrow \\\\\frac{dz}{dx}=\frac{1}{2}\sec^2\left(\frac{x}{2}\right)\longrightarrow \sec^2(x)=\tan^2(x)+1\implies \frac{dz}{dx}=\frac{z^2+1}{2}\\dz=\frac{1+z^2}{2}dx\implies dx=\frac{2}{1+z^2}dz\\\\ \sin(x)=\frac{2z}{1+z^2}~~~~~~~~ \tan(x)=\frac{2z}{1-z^2}\\\\\\ \int\frac{1}{\sin x+\tan x}dx=\int\frac{1}{\frac{2z}{1+z^2}+\frac{2z}{1-z^2}}\cdot\frac{2}{1+z^2}dz$

$\displaystyle \int\frac{1}{\frac{2z}{1+z^2}+\frac{2z}{1-z^2}}\cdot\frac{2}{1+z^2}dz=\int-\frac{z^2-1}{2z}dz=-\int\frac{z^2-1}{2z}dz \implies\\\\\\ -\frac{1}{2}\int\frac{z^2}{z}-\frac{1}{z}dz=-\frac{1}{2}\int z-\frac{1}{z}dz=-\frac{1}{2}\left(\frac{1}{2}z^2-\ln|z|\right)\implies \\\\ -\frac{1}{4}z^2+\frac{1}{2}\ln|z|\\\\ z=\tan\left(\frac{x}{2}\right)\\\\ \int\frac{1}{\sin(x)+\tan(x)}dx=\boxed{\frac{1}{2}\ln\left|\tan\left(\frac{x}{2}\right) \right|-\frac{1}{4}\tan^2\left(\frac{x}{2}\right)+C}$

Answer 2

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Calcular a integral indefinida:

$\large\begin{array}{l}\displaystyle\int\!\dfrac{1}{\mathrm{sen\,}x+\mathrm{tg\,}x}\,dx\\\\ =\displaystyle\int\!\dfrac{1}{\mathrm{sen\,}x+\frac{\mathrm{sen\,}x}{\cos x}}\,dx\\\\ =\displaystyle\int\!\dfrac{\cos x}{\mathrm{sen\,}x\cos x+\mathrm{sen\,}x}\,dx\\\\ =\displaystyle\int\!\dfrac{\cos x}{\mathrm{sen\,}x\cdot (\cos x+1)}\,dx\qquad\quad\mathbf{(i)} \end{array}$



Substituição:

$\large\begin{array}{l}z=\mathrm{tg\,}\dfrac{x}{2}\qquad\left(-\,\dfrac{\pi}{4}<x<\dfrac{\pi}{4} \right )\end{array}$



Daqui, segue que

$\large\begin{array}{l}\dfrac{x}{2}=\mathrm{arctg\,}z\quad\Rightarrow\quad \left\{\! \begin{array}{rcl} x&\!\!\!=\!\!\!&2\,\mathrm{arctg\,}z\\\\ dx&\!\!\!=\!\!\!&\dfrac{2}{1+z^2}\,dz \end{array} \right.\end{array}$



Também temos

$\large\begin{array}{l}z=\dfrac{\mathrm{sen\,}\frac{x}{2}}{\cos \frac{x}{2}}\\\\ z=\dfrac{2\,\mathrm{sen\,}\frac{x}{2}}{2\cos \frac{x}{2}}\\\\ z=\dfrac{2\,\mathrm{sen\,}\frac{x}{2}}{2\cos \frac{x}{2}}\cdot \dfrac{\cos\frac{x}{2}}{\cos\,\frac{x}{2}}\\\\ z=\dfrac{2\,\mathrm{sen\,}\frac{x}{2}\cos \frac{x}{2}}{2\cos\frac{x}{2}\cos\,\frac{x}{2}}\end{array}$

$\large\begin{array}{l}z=\dfrac{2\,\mathrm{sen\,}\frac{x}{2}\cos \frac{x}{2}}{2\cos^2\frac{x}{2}}\\\\ z=\dfrac{2\,\mathrm{sen\,}\frac{x}{2}\cos \frac{x}{2}}{1+\big(2\cos^2\frac{x}{2}-1\big)}\\\\\\ \therefore~~z=\dfrac{\mathrm{sen}\,x}{1+\cos x}\qquad\quad\mathbf{(ii)}\end{array}$



Expressando $\mathrm{sen\,}x$ e $\cos x$ em função de $z:$

•   $\large\begin{array}{l}\mathrm{sen\,}x\end{array}$

$\large\begin{array}{l}=2\,\mathrm{sen\,}\dfrac{x}{2}\cos\dfrac{x}{2}\\\\ =\dfrac{2\,\mathrm{sen\,}\frac{x}{2}\cos \frac{x}{2}}{1}\\\\ =\dfrac{2\,\mathrm{sen\,}\frac{x}{2}\cos \frac{x}{2}}{\cos^2 \frac{x}{2}+\mathrm{sen^2\,} \frac{x}{2}}\end{array}$

$\large\begin{array}{l}=\dfrac{2\,\mathrm{sen\,}\frac{x}{2}\cos \frac{x}{2}\cdot \frac{1}{\cos^2 \frac{x}{2}}}{\left(\cos^2 \frac{x}{2}+\mathrm{sen^2\,} \frac{x}{2}\right)\cdot \frac{1}{\cos^2 \frac{x}{2}}}\\\\ =\dfrac{2\,\frac{\mathrm{sen\,}\frac{x}{2}}{\cos \frac{x}{2}}}{1+\frac{\mathrm{sen^2\,} \frac{x}{2}}{\cos^2 \frac{x}{2}}}\\\\ =\dfrac{2\,\mathrm{tg\,}\frac{x}{2}}{1+\mathrm{tg^2\,}\frac{x}{2}}\\\\\\ \therefore~~\mathrm{sen\,}x=\dfrac{2z}{1+z^2}\qquad\quad\mathbf{(iii)}\end{array}$



De forma análoga,

•   $\large\begin{array}{l}\cos x\end{array}$

$\large\begin{array}{l}=\cos^2 \dfrac{x}{2}-\mathrm{sen^2\,}\dfrac{x}{2}\\\\ =\dfrac{\cos^2 \frac{x}{2}-\mathrm{sen^2\,}\frac{x}{2}}{1}\\\\ =\dfrac{\cos^2 \frac{x}{2}-\mathrm{sen^2\,}\frac{x}{2}}{\cos^2 \frac{x}{2}+\mathrm{sen^2\,}\frac{x}{2}}\\\\ =\dfrac{\left(\cos^2 \frac{x}{2}-\mathrm{sen^2\,}\frac{x}{2}\right)\cdot \frac{1}{\cos^2\frac{x}{2}}}{\left(\cos^2 \frac{x}{2}+\mathrm{sen^2\,}\frac{x}{2}\right)\cdot \frac{1}{\cos^2\frac{x}{2}}}\end{array}$

$\large\begin{array}{l}=\dfrac{1-\mathrm{tg^2\,}\frac{x}{2}}{1+\mathrm{tg\,}^2\frac{x}{2}}\\\\\\ \therefore~~\cos x=\dfrac{1-z^2}{1+z^2}\qquad\quad\mathbf{(iv)}\end{array}$

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Fazendo a substituição, a integral $\mathbf{(i)}$ fica

$\large\begin{array}{l}=\displaystyle\int\!\dfrac{\frac{1-z^2}{1+z^2}}{\frac{2z}{1+z^2}\cdot \left(\frac{1-z^2}{1+z^2}+1 \right )}\cdot \frac{2}{1+z^2}\,dz\\\\ =\displaystyle\int\!\dfrac{\frac{1-z^2}{1+z^2}}{\diagup \hspace{-7} 2z\cdot \left(\frac{1-z^2}{1+z^2}+1 \right )}\cdot \diagup \hspace{-7} 2\,dz\\\\ =\displaystyle\int\!\dfrac{1-z^2}{z\cdot \left(\frac{1-z^2}{1+z^2}+1 \right )\cdot (1+z^2)}\,dz \end{array}$

$\large\begin{array}{l}=\displaystyle\int\!\dfrac{1-z^2}{z\cdot (1-\diagup \hspace{-9} z^2+1+\diagup \hspace{-9}z^2)}\,dz\\\\ =\displaystyle\int\!\dfrac{1-z^2}{2z}\,dz\\\\ =\displaystyle\int\!\dfrac{1}{2z}\,dz-\int\!\dfrac{z^2}{2z}\,dz\\\\ =\displaystyle\frac{1}{2}\int\!\dfrac{1}{z}\,dz-\frac{1}{2}\int\!z\,dz \end{array}$

$\large\begin{array}{l}=\dfrac{1}{2}\,\mathrm{\ell n}|z|-\dfrac{1}{2}\cdot \dfrac{z^2}{2}+C\\\\ =\dfrac{1}{2}\,\mathrm{\ell n}|z|-\dfrac{1}{4}\,z^2+C\\\\ =\dfrac{1}{2}\,\mathrm{\ell n}\left|\dfrac{\mathrm{sen\,}x}{1+\cos x}\right|-\dfrac{1}{4}\cdot \left(\dfrac{\mathrm{sen\,}x}{1+\cos x}\right)^{\!\!2}+C\\\\ \end{array}$


$\large\boxed{\begin{array}{l}\displaystyle\int\!\frac{1}{\mathrm{sen\,}x+\mathrm{tg\,}x}=\frac{1}{2}\,\mathrm{\ell n}\left|\frac{\mathrm{sen\,}x}{1+\cos x}\right|-\frac{1}{4}\cdot \left(\frac{\mathrm{sen\,}x}{1+\cos x}\right)^{\!\!2}+C \end{array}}\qquad\checkmark$



Bons estudos! :-)


Tags:   integral indefinida trigonométrica substituição tangente tg tan x/2 sen sen sin cosseno cos função racional frações parciais cálculo integral