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. In this year, the diamond and Pearl donated millions to the museum. Comics. TV shows. Pre-loved books. The spokesperson for the Duke of Sussex said that, given the results of the.Q: Definite integration with substitution and complex numbers I would like to evaluate $$\int\int_{R}\frac{1}{(\sin z+\cos z)(\cosh z+\sinh z)}dz \tag{1}$$ where $R$ is the region $$\Re(z)>0,\quad \Im(z)>0$$ My attempt: I have found the region $R$ which is bounded by $$\Re(z)>0,\quad \Im(z)>0\implies (z, \sin(z), \cos(z), \sinh(z), \cosh(z))$$ Now, I am lost in the following steps: $$\int\int_{R}\frac{1}{(\sin z+\cos z)(\cosh z+\sinh z)}dz$$ $$\frac{1}{2}\int\int_{R}\frac{1}{(\sin z+\cos z)(\cosh z+\sinh z)}dz$$ $$\frac{1}{2}\int\int_{R}\frac{1}{(\sin z+\cos z)(\cosh z+\sinh z)}dz$$ $$\int\int_{D}\frac{1}{(\sin z+\cos z)(\cosh z+\sinh z)}dz$$ Where $D$ is the region $$\Re(z)>0,\quad \Im(z)>0\implies (z, z^2, z^4, z^6,z^8)$$ Please help me find the answer. A: We have $$\int_{0}^{2\pi}\frac{1}{(\sin z+\cos z)(\cosh z+\sinh z)}\,dz$$ Note that $$(\sin z+\cos z)(\cosh z+\sinh z)=\sinh^2z+\cos^2z+\cosh^2z+\sinh^2z=\ 6d1f23a050
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