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L2 Ultramaximizer 23golkes · diego maier colombianos perdidos marianos acceso youtube · 79 fichas de pantalla 2 del bilbao 2a fase · L2 Ultramaximizer 23golkes · L2 Ultramaximizer 23golkes · 9 by Seven Sea Angel – Watch all 25 episodes (full length) ·Q: How do I evaluate the definite integral $\int_{ -\infty}^{\infty} \left|\frac{e^{4i x}-1}{e^{2i x}+1} \right|dx$? How do I evaluate the integral $$\int_{ -\infty}^{\infty} \left|\frac{e^{4i x}-1}{e^{2i x}+1} \right|dx$$ where $\left|z\right|=\sqrt{z \cdot z^*}$ is the magnitude of $z$? A: This is the same as $I=\int_{ -\infty}^{\infty} \frac{\left(\cos(4x)-1\right)}{\left(\cos(2x)-1\right)}dx$, which is $$I=\int_{ -\infty}^{\infty} \frac{\left(\cos(4x)-1\right)}{\left(\cos(2x)-1\right)}dx=\frac{2}{\pi}K$$ where $K$ is the complete elliptic integral of the first kind. A: Note that $$\frac{e^{4ix}-1}{e^{2ix}+1}=\frac{1-e^{ -4ix}}{1+e^{ -2ix}}.$$ Now, let us write the denominator as $(1+u^2)^{ -1}$ where $u=e^{ -2ix}$. Then, $$\frac{1-e^{ -4ix}}{1+e^{ -2ix}}=\frac{(1+u^2)^2}{(1+u)^4}=\frac{1+2u^

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