Question Number 54756 by Knight last updated on 10/Feb/19
$${solve} \\ $$$$\int{tan}\left({x}−\theta\right){tan}\left({x}+\theta\right){tan}\:\mathrm{2}{x}\:{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19
![tan2x=tan(x+θ+x−θ) tan2x=((tan(x+θ)+tan(x−θ))/(1−tan(x+θ)tan(x−θ))) tan2x−tan2xtan(x+θ)tan(x−θ)=tan(x+θ)+tan(x−θ) tan2x−tan(x+θ)−tan(x−θ)=tan2xtan(x+θ)tan(x−θ) so ∫tan(x−θ)tan(x+θ)tan2xdx =∫[tan2x−tan(x+θ)−tan(x−θ)]dx ∫tan2xdx−∫tan(x+θ)dx−∫tan(x−θ)dx =((ln(sec2x))/2)−((lnsec(x+θ))/1)−((lnsec(x−θ))/1)+c](https://www.tinkutara.com/question/Q54757.png)
$${tan}\mathrm{2}{x}={tan}\left({x}+\theta+{x}−\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{tan}\mathrm{2}{x}=\frac{{tan}\left({x}+\theta\right)+{tan}\left({x}−\theta\right)}{\mathrm{1}−{tan}\left({x}+\theta\right){tan}\left({x}−\theta\right)} \\ $$$${tan}\mathrm{2}{x}−{tan}\mathrm{2}{xtan}\left({x}+\theta\right){tan}\left({x}−\theta\right)={tan}\left({x}+\theta\right)+{tan}\left({x}−\theta\right) \\ $$$${tan}\mathrm{2}{x}−{tan}\left({x}+\theta\right)−{tan}\left({x}−\theta\right)={tan}\mathrm{2}{xtan}\left({x}+\theta\right){tan}\left({x}−\theta\right) \\ $$$${so}\:\int{tan}\left({x}−\theta\right){tan}\left({x}+\theta\right){tan}\mathrm{2}{xdx} \\ $$$$=\int\left[{tan}\mathrm{2}{x}−{tan}\left({x}+\theta\right)−{tan}\left({x}−\theta\right)\right]{dx} \\ $$$$\int{tan}\mathrm{2}{xdx}−\int{tan}\left({x}+\theta\right){dx}−\int{tan}\left({x}−\theta\right){dx} \\ $$$$=\frac{{ln}\left({sec}\mathrm{2}{x}\right)}{\mathrm{2}}−\frac{{lnsec}\left({x}+\theta\right)}{\mathrm{1}}−\frac{{lnsec}\left({x}−\theta\right)}{\mathrm{1}}+{c} \\ $$