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tan-2x-3-dx-Mastermind-




Question Number 169362 by Mastermind last updated on 29/Apr/22
∫tan(2x+3)dx    Mastermind
$$\int{tan}\left(\mathrm{2}{x}+\mathrm{3}\right){dx} \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by peter frank last updated on 29/Apr/22
let u=2x+3  du=2dx  ∫tan u.(du/2)
$$\mathrm{let}\:\mathrm{u}=\mathrm{2x}+\mathrm{3} \\ $$$$\mathrm{du}=\mathrm{2dx} \\ $$$$\int\mathrm{tan}\:\mathrm{u}.\frac{\mathrm{du}}{\mathrm{2}} \\ $$
Commented by Mastermind last updated on 29/Apr/22
thanks, you did not complete it tho.
$${thanks},\:{you}\:{did}\:{not}\:{complete}\:{it}\:{tho}. \\ $$
Answered by Mathspace last updated on 29/Apr/22
∫ tan(2x+3)dx=_(2x+3=t) ∫ tant (dt/2)  =(1/2)∫ ((sint)/(cost))dt =−(1/2)ln∣cost∣ +c  =−(1/2)ln∣cos(2x+3)∣ +c
$$\int\:{tan}\left(\mathrm{2}{x}+\mathrm{3}\right){dx}=_{\mathrm{2}{x}+\mathrm{3}={t}} \int\:{tant}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{sint}}{{cost}}{dt}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{cost}\mid\:+{c} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{cos}\left(\mathrm{2}{x}+\mathrm{3}\right)\mid\:+{c} \\ $$

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