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l-sin-7xdx-




Question Number 217408 by Intesar last updated on 13/Mar/25
l∫sin 7xdx
$${l}\int\mathrm{sin}\:\mathrm{7}{xdx} \\ $$
Answered by SdC355 last updated on 13/Mar/25
−(1/7)cos(7x)+C  because.  ((d  )/dt) cos(t)=−sin(t)  ∫  sin(at)dt=− (1/a)cos(at)+C  , a∈C\{0}  C is Field of Complex number  if a=0   (1/0) ∙cos(0t)+C    1/0 is divergence as ±∞
$$−\frac{\mathrm{1}}{\mathrm{7}}\mathrm{cos}\left(\mathrm{7}{x}\right)+{C} \\ $$$$\mathrm{because}. \\ $$$$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\:\mathrm{cos}\left({t}\right)=−\mathrm{sin}\left({t}\right) \\ $$$$\int\:\:\mathrm{sin}\left({at}\right)\mathrm{d}{t}=−\:\frac{\mathrm{1}}{{a}}\mathrm{cos}\left({at}\right)+{C}\:\:,\:{a}\in\mathbb{C}\backslash\left\{\mathrm{0}\right\} \\ $$$$\mathbb{C}\:\mathrm{is}\:\mathrm{Field}\:\mathrm{of}\:\mathrm{Complex}\:\mathrm{number} \\ $$$$\mathrm{if}\:\mathrm{a}=\mathrm{0}\:\:\:\frac{\mathrm{1}}{\mathrm{0}}\:\centerdot\mathrm{cos}\left(\mathrm{0}{t}\right)+{C}\:\: \\ $$$$\mathrm{1}/\mathrm{0}\:\mathrm{is}\:\mathrm{divergence}\:\mathrm{as}\:\pm\infty \\ $$

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