## Calculus: Early Transcendentals 8th Edition

$$\int^{10}_{-10}\frac{2e^x}{\sinh x+\cosh x}dx=40$$
$$A=\int^{10}_{-10}\frac{2e^x}{\sinh x+\cosh x}dx$$ $$A=\int^{10}_{-10}Bdx$$ *Consider B: $$B=\frac{2e^x}{\sinh x+\cosh x}$$ We know that $$\sinh x=\frac{e^x-e^{-x}}{2}$$ $$\cosh x=\frac{e^x+e^{-x}}{2}$$ Therefore, $$B=\frac{2e^x}{\frac{e^x-e^{-x}}{2}+\frac{e^x+e^{-x}}{2}}$$ $$B=\frac{2e^x}{\frac{2e^x}{2}}$$ $$B=\frac{2e^x}{e^x}$$ $$B=2$$ That means $$A=\int^{10}_{-10}2dx$$ According to Table 1, we have $$\int kdx=kx+C$$ Therefore, $$A=2x\Bigg]^{10}_{-10}$$ $$A=2[10-(-10)]$$ $$A=40$$