Question

alguém pode me ajudar? integral de 3dx/8+7cosx
pela técnica de substituição z=tgx/2

Answer 1

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Calcular a integral indefinida:

$\large\begin{array}{l}\displaystyle\int\!\dfrac{3}{8+7\cos x}\,dx\end{array}$



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

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

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



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

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

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

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



De forma análoga,

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

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

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

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Fazendo a substituição, a integral fica

$\large\begin{array}{l}=\displaystyle\int\!\dfrac{3}{8+7\,\frac{1-z^2}{1+z^2}}\cdot \dfrac{2}{1+z^2}\,dz\\\\\\ =\displaystyle\int\!\dfrac{6}{8(1+z^2)+7(1-z^2)}\,dz\\\\\\ =\displaystyle\int\!\dfrac{6}{8+8z^2+7-7z^2}\,dz\\\\\\ =\displaystyle\int\!\dfrac{6}{z^2+15}\,dz\\\\\\ =\displaystyle\int\!\dfrac{6}{z^2+\big(\sqrt{15}\big)^2}\,dz\end{array}$

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Temos tabelada esta integral:

$\large\begin{array}{l}\displaystyle\int\!\dfrac{1}{z^2+a^2}=\dfrac{1}{a}\,\mathrm{arctg\,}\dfrac{z}{a}+C\qquad\quad\mathbf{(iv)}\end{array}$

(que pode ser facilmente obtida por substituição trigonométrica)

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Então, a integral fica

$\large\begin{array}{l}=\dfrac{6}{\sqrt{15}}\,\mathrm{arctg\,}\dfrac{z}{\sqrt{15}}+C\\\\\\ =\dfrac{6}{\sqrt{15}}\,\mathrm{arctg}\!\left[\dfrac{~\frac{\mathrm{sen\,}x}{\cos x+1}~}{\sqrt{15}}\right]+C\\\\\\ =\dfrac{6}{\sqrt{15}}\,\mathrm{arctg}\!\left[\dfrac{\mathrm{sen\,}x}{\sqrt{15}\cdot (\cos x+1)}\right]+C\end{array}$


$\large\begin{array}{l}\therefore~~\boxed{\begin{array}{c}\displaystyle\int\!\dfrac{3}{8+7\cos x}\,dx=\frac{6}{\sqrt{15}}\,\mathrm{arctg}\!\left[\dfrac{\mathrm{sen\,}x}{\sqrt{15}\cdot (\cos x+1)}\right]+C \end{array}}\qquad\quad\checkmark\end{array}$



Bons estudos! :-)


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