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Question Number 71499 by petrochengula last updated on 16/Oct/19

Commented by petrochengula last updated on 16/Oct/19

$$\mathrm{help}2\mathrm{a}$$

Answered by MJS last updated on 16/Oct/19

$$\cosh x=\frac{1}{2}({\mathrm{e}}^x+{\mathrm{e}}^{-x})$$$$\cosh 2x=\frac{1}{2}({\mathrm{e}}^{2x}+{\mathrm{e}}^{-2x})$$$$\frac{\cosh 2x}{\cosh x}={\mathrm{e}}^x+{\mathrm{e}}^{-x}-\frac{2}{{\mathrm{e}}^x+{\mathrm{e}}^{-x}}=2\cosh x-\frac{1}{\cosh x}\Rightarrow$$$$\Rightarrow \cosh 2x={2\cosh }^{2}x-1$$$$\cosh 3x=\frac{1}{2}({\mathrm{e}}^{3x}+{\mathrm{e}}^{-3x})$$$$\frac{\cosh 3x}{\cosh x}={\mathrm{e}}^{2x}+{\mathrm{e}}^{-2x}-1=2\cosh 2x-1=$$$$={4\cosh }^{2}x-3$$$$\Rightarrow \cosh 3x={4\cosh }^{3}x-3\cosh x$$$$$$$$a\cosh 3x+b\cosh 2x=0$$$$[\cosh x=c]$$$$4{ac}^3+2{bc}^2-3ac-b=0$$$$c^3+\frac{b}{2a}{c}^2-\frac{3}{4}c-\frac{b}{4a}=0$$$$(c-{\gamma }_1)(c-{\gamma }_2)(c-{\gamma }_3)=$$$$=c^3-({\gamma }_1+{\gamma }_2+{\gamma }_3)c^2+({\gamma }_1{\gamma }_2+{\gamma }_1{\gamma }_3+{\gamma }_2{\gamma }_3)c-{\gamma }_1{\gamma }_2{\gamma }_3$$$$\Rightarrow {\gamma }_1{\gamma }_2+{\gamma }_1{\gamma }_3+{\gamma }_2{\gamma }_3=-\frac{3}{4}$$$$\mathrm{independent}\mathrm{of}a,b$$

Commented by petrochengula last updated on 16/Oct/19

$$\mathrm{thanks}$$$$$$

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