Question Number 1597 by Rasheed Ahmad last updated on 25/Aug/15
$$\int\frac{{f}\left({x}\right)}{{f}\:'\left({x}\right)}{dx}={ln}\:{sec}\:{x}+{c} \\ $$$${f}\left({x}\right)=? \\ $$
Answered by 123456 last updated on 25/Aug/15
![∫(y/(dy/dx))dx=ln sec x+c (d/dx)[∫(y/(dy/dx))dx]=(d/dx)[ln sec x+c] (y/(dy/dx))=((sec x tan x)/(sec x))=tan x y=(dy/dx)tan x (dy/y)=(dx/(tan x)) ∫(dy/y)=∫cot xdx ln y=∫cot xdx=ln sin x+g y=e^(ln sin x+g) =e^(ln sin x) e^g y=ksin x,k=e^g −−−−−−−−−−−−−−−−−−−− verification y=ksin x dy/dx=kcos x ∫(y/(dy/dx))dx=∫((ksin x)/(kcos x))dx =∫((sin x)/(cos x))dx,u=cos x,du=−sin xdx =−∫(du/u) =−ln u+C =−ln cos x+C =ln (cos x)^(−1) +C=ln (1/(cos x))+C =ln sec x+C](https://www.tinkutara.com/question/Q1598.png)
$$\int\frac{{y}}{{dy}/{dx}}{dx}=\mathrm{ln}\:\mathrm{sec}\:{x}+{c} \\ $$$$\frac{{d}}{{dx}}\left[\int\frac{{y}}{{dy}/{dx}}{dx}\right]=\frac{{d}}{{dx}}\left[\mathrm{ln}\:\mathrm{sec}\:{x}+{c}\right] \\ $$$$\frac{{y}}{{dy}/{dx}}=\frac{\mathrm{sec}\:{x}\:\mathrm{tan}\:{x}}{\mathrm{sec}\:{x}}=\mathrm{tan}\:{x} \\ $$$${y}=\frac{{dy}}{{dx}}\mathrm{tan}\:{x} \\ $$$$\frac{{dy}}{{y}}=\frac{{dx}}{\mathrm{tan}\:{x}} \\ $$$$\int\frac{{dy}}{{y}}=\int\mathrm{cot}\:{xdx} \\ $$$$\mathrm{ln}\:{y}=\int\mathrm{cot}\:{xdx}=\mathrm{ln}\:\mathrm{sin}\:{x}+{g} \\ $$$${y}={e}^{\mathrm{ln}\:\mathrm{sin}\:{x}+{g}} ={e}^{\mathrm{ln}\:\mathrm{sin}\:{x}} {e}^{{g}} \\ $$$${y}={k}\mathrm{sin}\:{x},{k}={e}^{{g}} \\ $$$$−−−−−−−−−−−−−−−−−−−− \\ $$$$\mathrm{verification} \\ $$$${y}={k}\mathrm{sin}\:{x} \\ $$$${dy}/{dx}={k}\mathrm{cos}\:{x} \\ $$$$\int\frac{{y}}{{dy}/{dx}}{dx}=\int\frac{{k}\mathrm{sin}\:{x}}{{k}\mathrm{cos}\:{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}{dx},{u}=\mathrm{cos}\:{x},{du}=−\mathrm{sin}\:{xdx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\int\frac{{du}}{{u}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{ln}\:{u}+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{ln}\:\mathrm{cos}\:{x}+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{ln}\:\left(\mathrm{cos}\:{x}\right)^{−\mathrm{1}} +\mathrm{C}=\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{cos}\:{x}}+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{ln}\:\mathrm{sec}\:{x}+\mathrm{C} \\ $$
Commented by 123456 last updated on 25/Aug/15
$$\frac{{d}\left(\mathrm{sec}\:{x}\right)}{{dx}}=\frac{{d}}{{dx}}\left(\frac{\mathrm{1}}{\mathrm{cos}\:{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\frac{{d}\left(\mathrm{cos}\:{x}\right)}{{dx}}}{\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\centerdot\frac{\mathrm{1}}{\mathrm{cos}\:{x}}=\mathrm{tan}\:{x}\:\mathrm{sec}\:{x} \\ $$$$\frac{{d}}{{dx}}\mathrm{ln}\:{u}=\frac{{du}/{dx}}{{u}} \\ $$$$\int\mathrm{cot}\:{xdx}=\int\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}{dx}=\int\frac{{du}}{{u}}=\mathrm{ln}\:{u}+\mathrm{C}=\mathrm{ln}\:\mathrm{sin}\:{x}+\mathrm{C} \\ $$$${u}=\mathrm{sin}\:{x},{du}=\mathrm{cos}\:{xdx} \\ $$
Commented by Rasheed Ahmad last updated on 25/Aug/15
$$\mathrm{Excellent}! \\ $$