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Question Number 115366 by Bird last updated on 25/Sep/20

$$calculate{\int }_{-1}^2\frac{dx}{{ch}^{2}x+{sh}^{2}x}$$

Answered by MJS_new last updated on 25/Sep/20

$$\int \frac{dx}{{\cosh }^{2}x+{\sinh }^{2}x}=$$$$[t={\mathrm{e}}^{2x}\to dx=\frac{dt}{{2\mathrm{e}}^{2x}}]$$$$=\int \frac{dt}{t^2+1}=\arctan t=$$$$=\arctan {\mathrm{e}}^{2x}+C$$$$\underset{-1}{\stackrel{2}{\int }}\frac{dx}{{\cosh }^{2}x+{\sinh }^{2}x}=\arctan {\mathrm{e}}^4-\arctan {\mathrm{e}}^{-2}$$

Commented by mathmax by abdo last updated on 25/Sep/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}\mathrm{mjs}$$

Answered by mathmax by abdo last updated on 25/Sep/20

$$\mathrm{let}\mathrm{I}={\int }_{-1}^2\frac{\mathrm{dx}}{{\mathrm{ch}}^2\mathrm{x}+{\mathrm{sh}}^2\mathrm{x}}\Rightarrow \mathrm{I}={\int }_{-1}^2\frac{\mathrm{dx}}{\frac{1+\mathrm{ch}\left(2\mathrm{x}\right)}{2}+\frac{\mathrm{ch}\left(2\mathrm{x}\right)-1}{2}}$$$$={\int }_{-1}^2\frac{\mathrm{dx}}{\mathrm{ch}\left(2\mathrm{x}\right)}=2{\int }_{-1}^2\frac{\mathrm{dx}}{{\mathrm{e}}^{2\mathrm{x}}+{\mathrm{e}}^{-2\mathrm{x}}}{=}_{{\mathrm{e}}^{2\mathrm{x}}=\mathrm{t}}2{\int }_{{\mathrm{e}}^{-2}}^{{\mathrm{e}}^{-4}}\frac{1}{\mathrm{t}+{\mathrm{t}}^{-1}}\times \frac{\mathrm{dt}}{2\mathrm{t}}$$$$={\int }_{{\mathrm{e}}^{-2}}^{{\mathrm{e}}^{-4}}\frac{\mathrm{dt}}{{\mathrm{t}}^2+1}=\arctan \left(\frac{1}{{\mathrm{e}}^4}\right)-\arctan \left(\frac{1}{{\mathrm{e}}^2}\right)$$$$=\frac{\pi }{2}-\arctan \left({\mathrm{e}}^4\right)-(\frac{\pi }{2}-\arctan \left({\mathrm{e}}^2\right))=\arctan \left({\mathrm{e}}^2\right)-\arctan \left({\mathrm{e}}^4\right)$$$$$$

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