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Question Number 178996 by Spillover last updated on 23/Oct/22

Answered by qaz last updated on 23/Oct/22

$$\left(a\right).$$$$2sinhx-cosh=(e^x-e^{-x})-2(e^x+e^{-x})$$$$=-e^x-3e^{-x}$$$$2tanh\frac{x}{2}=2\cdot \frac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{e^{\frac{x}{2}}+e^{-\frac{x}{2}}}=\frac{2e^x-2}{e^x+1}$$$$\Rightarrow -e^x-3e^{-x}=\frac{2e^x-2}{e^x+1}$$$$\Rightarrow e^{3x}+3e^{2x}+e^x+3=(e^x+3)(e^{2x}+1)=0$$$$\Rightarrow x=(\frac{1}{2}+k)i\pi orx=ln3+(2k+1)i\pi ,k\in Z$$

Answered by Frix last updated on 23/Oct/22

$$2\sinh x-\cosh x=2\tanh \frac{x}{2}$$$${\mathrm{e}}^x-{\mathrm{e}}^{-x}-\frac{{\mathrm{e}}^x+{\mathrm{e}}^{-x}}{2}=2\frac{{\mathrm{e}}^x-1}{{\mathrm{e}}^x+1}$$$${\mathrm{e}}^{3x}-{3\mathrm{e}}^{2x}+{\mathrm{e}}^x-3=0$$$$({\mathrm{e}}^x-3)({\mathrm{e}}^{2x}+1)=0$$$$x=\ln 3$$$$$$$${16\sinh }^{2}x{\cosh }^{2}x=$$$$=16{\left(\frac{{\mathrm{e}}^x-{\mathrm{e}}^{-x}}{2}\right)}^2{\left(\frac{{\mathrm{e}}^x+{\mathrm{e}}^{-x}}{2}\right)}^3=$$$$=\frac{{\mathrm{e}}^{5x}+{\mathrm{e}}^{-5x}+{\mathrm{e}}^{3x}+{\mathrm{e}}^{-3x}-{2\mathrm{e}}^x-{2\mathrm{e}}^{-x}}{2}=$$$$=\cosh 5x+\cosh 3x-2\cosh x$$$$$$$$16\underset{0}{\stackrel{1}{\int }}{\sinh }^{2}x{\cosh }^{2}xdx\stackrel{t={\mathrm{e}}^x}{=}$$$$=\underset{1}{\stackrel{\mathrm{e}}{\int }}(\frac{t^4}{2}+\frac{t^2}{2}-1-\frac{1}{t^2}+\frac{1}{2t^4}+\frac{1}{2t^6})dt=$$$$={[\frac{t^5}{10}+\frac{t^3}{6}-t+\frac{1}{t}-\frac{1}{6t^3}-\frac{1}{10t^5}]}_1^{\mathrm{e}}=$$$$=\frac{{\mathrm{e}}^5}{10}+\frac{{\mathrm{e}}^3}{6}-\mathrm{e}+\frac{1}{t\mathrm{le}}-\frac{1}{{6\mathrm{e}}^3}-\frac{1}{{10\mathrm{e}}^5}=$$$$=\frac{\sinh 5}{5}+\frac{\sinh 3}{3}-2\sinh 1$$

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