Question Number 140581 by bounhome last updated on 09/May/21
$${prove}\:\:\int{sec}^{\mathrm{2}} {xdx}={tanx} \\ $$
Answered by MJS_new last updated on 09/May/21
![tan x =((sin x)/(cos x)) (d/dx)[((u(x))/(v(x)))]=((u′(x)v(x)−u(x)v′(x))/((v(x))^2 )) ⇒ (d/dx)[((sin x)/(cos x))]=((cos x cos x −sin x (−sin x))/(cos^2 x))= =((cos^2 x +sin^2 x)/(cos^2 x))=(1/(cos^2 x))=sec^2 x ∫((d/dx)[f(x)])dx=f(x)+C ⇒ ∫sec^2 x dx=tan x +C](https://www.tinkutara.com/question/Q140585.png)
$$\mathrm{tan}\:{x}\:=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$\frac{{d}}{{dx}}\left[\frac{{u}\left({x}\right)}{{v}\left({x}\right)}\right]=\frac{{u}'\left({x}\right){v}\left({x}\right)−{u}\left({x}\right){v}'\left({x}\right)}{\left({v}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\frac{{d}}{{dx}}\left[\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right]=\frac{\mathrm{cos}\:{x}\:\mathrm{cos}\:{x}\:−{s}\mathrm{in}\:{x}\:\left(−\mathrm{sin}\:{x}\right)}{\mathrm{cos}^{\mathrm{2}} \:{x}}= \\ $$$$=\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\mathrm{sec}^{\mathrm{2}} \:{x} \\ $$$$\int\left(\frac{{d}}{{dx}}\left[{f}\left({x}\right)\right]\right){dx}={f}\left({x}\right)+{C} \\ $$$$\Rightarrow \\ $$$$\int\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx}=\mathrm{tan}\:{x}\:+{C} \\ $$
Answered by MJS_new last updated on 10/May/21
![∫sec^2 x dx= [t=tan x → dx=cos^2 x dt] =∫dt=t=tan x +C](https://www.tinkutara.com/question/Q140591.png)
$$\int\mathrm{sec}^{\mathrm{2}} \:{x}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}\right] \\ $$$$=\int{dt}={t}=\mathrm{tan}\:{x}\:+{C} \\ $$