Question Number 170311 by Mastermind last updated on 20/May/22

$$\mathrm{5}\frac{{dy}}{{dx}}={tan}\left(\mathrm{2}{x}+\mathrm{2}{y}+\mathrm{6}\right) \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by mr W last updated on 20/May/22
![t=2x+2y+6 (dt/dx)=2+2(dy/dx) 5((dt/dx)−2)=2 tan t 5(dt/dx)=2(tan t+5) (dt/(tan t+5))=((2dx)/5) ∫(dt/(tan t+5))=∫((2dx)/5) ... ((5t+ln (sin t+5 cos t) )/(26))=((2x)/5)+C ((10(x+y+3)+ln [sin (2x+2y+6)+5 cos (2x+2y+6)] )/(26))=((2x)/5)+C](https://www.tinkutara.com/question/Q170314.png)
$${t}=\mathrm{2}{x}+\mathrm{2}{y}+\mathrm{6} \\ $$$$\frac{{dt}}{{dx}}=\mathrm{2}+\mathrm{2}\frac{{dy}}{{dx}} \\ $$$$\mathrm{5}\left(\frac{{dt}}{{dx}}−\mathrm{2}\right)=\mathrm{2}\:\mathrm{tan}\:{t} \\ $$$$\mathrm{5}\frac{{dt}}{{dx}}=\mathrm{2}\left(\mathrm{tan}\:{t}+\mathrm{5}\right) \\ $$$$\frac{{dt}}{\mathrm{tan}\:{t}+\mathrm{5}}=\frac{\mathrm{2}{dx}}{\mathrm{5}} \\ $$$$\int\frac{{dt}}{\mathrm{tan}\:{t}+\mathrm{5}}=\int\frac{\mathrm{2}{dx}}{\mathrm{5}} \\ $$$$… \\ $$$$\frac{\mathrm{5}{t}+\mathrm{ln}\:\left(\mathrm{sin}\:\:{t}+\mathrm{5}\:\mathrm{cos}\:{t}\right)\:}{\mathrm{26}}=\frac{\mathrm{2}{x}}{\mathrm{5}}+{C} \\ $$$$\frac{\mathrm{10}\left({x}+{y}+\mathrm{3}\right)+\mathrm{ln}\:\left[\mathrm{sin}\:\:\left(\mathrm{2}{x}+\mathrm{2}{y}+\mathrm{6}\right)+\mathrm{5}\:\mathrm{cos}\:\left(\mathrm{2}{x}+\mathrm{2}{y}+\mathrm{6}\right)\right]\:}{\mathrm{26}}=\frac{\mathrm{2}{x}}{\mathrm{5}}+{C} \\ $$
Commented by Mastermind last updated on 20/May/22

$${Sir},\:{Is}\:{that}\:{final}\:{anwer}? \\ $$$${and}\:{besides}\:{which}\:{method}\:{is}\:{this}? \\ $$
Commented by Mastermind last updated on 20/May/22

$${Thanks} \\ $$
Commented by Mastermind last updated on 20/May/22

$${I}\:{really}\:{appreciate}! \\ $$