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Question Number 146933 by mnjuly1970 last updated on 16/Jul/21

$$$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$$$$$\mathrm{I}:={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}sin(cosh\left(x\right).cos\left(sinhx\right))dx=?$$$$....solution....$$$$\mathrm{I}:=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\{sin(cosh\left(x\right)+sinh\left(x\right))+sin(cosh\left(x\right)-sinh\left(x\right))\}$$$$:\underset{sinh\left(x\right)=\frac{e^x-e^{-x}}{2}}{\stackrel{cosh\left(x\right)=\frac{e^x+e^{-x}}{2}}{=}}\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\{sin\left(e^x\right)+sin\left(e^{-x}\right)\}dx$$$$:=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}sin\left(e^x\right)dx+[\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}sin\left(e^x\right)dx::x\stackrel{sub}{=}-x]$$$$:={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}sin\left(e^x\right)dx\stackrel{e^x=y}{=}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt$$$$......\mathrm{I}:=\frac{\pi }{2}.....$$$$...m.n.1970...$$$$$$$$$$

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