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Question Number 99465 by mathmax by abdo last updated on 21/Jun/20

$$\mathrm{calculate}\mathrm{I}=\int {\cos }^2\mathrm{x}\mathrm{sh}\left(2\mathrm{x}\right)\mathrm{dx}\mathrm{and}\mathrm{J}=\int {\sin }^2\mathrm{x}\mathrm{ch}\left(2\mathrm{x}\right)\mathrm{dx}$$

Answered by mathmax by abdo last updated on 22/Jun/20

$$\mathrm{I}=\int {\cos }^2\mathrm{x}\mathrm{sh}\left(2\mathrm{x}\right)\mathrm{dx}\Rightarrow \mathrm{I}=\frac{1}{2}\int (1+\cos \left(2\mathrm{x}\right))\mathrm{sh}\left(2\mathrm{x}\right)\mathrm{dx}$$$$=\frac{1}{2}\int \mathrm{sh}\left(2\mathrm{x}\right)\mathrm{dx}+\frac{1}{2}\int \cos \left(2\mathrm{x}\right)\mathrm{sh}\left(2\mathrm{x}\right)\mathrm{dx}$$$$\int \mathrm{sh}\left(2\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}\mathrm{ch}\left(2\mathrm{x}\right)+{\mathrm{c}}_1$$$$\int \cos \left(2\mathrm{x}\right)\mathrm{sh}\left(2\mathrm{x}\right)\mathrm{dx}=\mathrm{Re}(\int {\mathrm{e}}^{2\mathrm{ix}}\times \frac{{\mathrm{e}}^{2\mathrm{ix}}-{\mathrm{e}}^{-2\mathrm{ix}}}{2}\mathrm{dx})$$$$\int \frac{{\mathrm{e}}^{4\mathrm{ix}}-1}{2}\mathrm{dx}=-\frac{\mathrm{x}}{2}+\frac{1}{8\mathrm{i}}{\mathrm{e}}^{4\mathrm{ix}}=-\frac{\mathrm{x}}{2}-\frac{\mathrm{i}}{8}(\cos \left(4\mathrm{x}\right)+\mathrm{isin}\left(4\mathrm{x}\right)\Rightarrow$$$$\mathrm{Re}(....)=-\frac{\mathrm{x}}{2}+\frac{1}{8}\sin \left(4\mathrm{x}\right)\Rightarrow \mathrm{I}=\frac{1}{2}\mathrm{ch}\left(2\mathrm{x}\right)-\frac{\mathrm{x}}{4}+\frac{1}{16}\sin \left(4\mathrm{x}\right)$$

Answered by abdomathmax last updated on 22/Jun/20

$$\mathrm{J}=\int {\sin }^2\mathrm{x}\mathrm{ch}\left(2\mathrm{x}\right)\mathrm{dx}\Rightarrow \mathrm{J}=\frac{1}{2}\int (1-\cos \left(2\mathrm{x}\right))\mathrm{ch}\left(2\mathrm{x}\right)\mathrm{dx}$$$$=\frac{1}{2}\int \mathrm{ch}\left(2\mathrm{x}\right)\mathrm{dx}-\frac{1}{2}\int \cos \left(2\mathrm{x}\right)\mathrm{ch}\left(2\mathrm{x}\right)\mathrm{dx}\mathrm{we}\mathrm{have}$$$$\int \mathrm{ch}\left(2\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}\mathrm{sh}\left(2\mathrm{x}\right)+{\mathrm{c}}_1$$$$\int \cos \left(2\mathrm{x}\right)\mathrm{ch}\left(2\mathrm{x}\right)\mathrm{dx}=\frac{1}{2}\mathrm{Re}(\int {\mathrm{e}}^{2\mathrm{ix}}({\mathrm{e}}^{2\mathrm{ix}}+{\mathrm{e}}^{-2\mathrm{ix}})\mathrm{dx})$$$$\mathrm{and}\int ({\mathrm{e}}^{4\mathrm{ix}}+1)\mathrm{dx}=\mathrm{x}+\frac{1}{4\mathrm{i}}{\mathrm{e}}^{4\mathrm{ix}}$$$$=\mathrm{x}-\frac{\mathrm{i}}{4}(\cos \left(4\mathrm{x}\right)+\mathrm{isin}\left(4\mathrm{x}\right))$$$$\Rightarrow \frac{1}{2}\mathrm{Re}(...)=\frac{\mathrm{x}}{2}+\frac{1}{8}\sin \left(4\mathrm{x}\right)\Rightarrow$$$$\mathrm{J}=\frac{1}{4}\mathrm{sh}\left(2\mathrm{x}\right)-\frac{\mathrm{x}}{5}-\frac{1}{16}\sin \left(4\mathrm{x}\right)+\mathrm{c}$$$$$$

Commented by abdomathmax last updated on 22/Jun/20

$$\mathrm{J}=\frac{1}{4}\mathrm{sh}\left(2\mathrm{x}\right)-\frac{\mathrm{x}}{4}-\frac{1}{16}\sin \left(4\mathrm{x}\right)+\mathrm{c}$$

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