Question Number 212099 by vahid last updated on 30/Sep/24
Answered by mehdee7396 last updated on 30/Sep/24
$$\int\frac{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}\:\:\:\:\:;\:{let}\:\:\:{tan}\frac{{x}}{\mathrm{2}}={u} \\ $$$$=\int\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }{du}={ln}\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}+{c} \\ $$$$={ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)\right)+{c}\:\:\checkmark \\ $$$$ \\ $$$$ \\ $$
Answered by Frix last updated on 30/Sep/24
![∫(dx/(cos x))=∫sec x dx= =∫sec x ((tan x +sec x)/(tan x +sec x))dx =^([t=tan x +sec x]) =∫(dt/t)=ln t = =ln ∣tan x +sec x∣ +C](https://www.tinkutara.com/question/Q212102.png)
$$\int\frac{{dx}}{\mathrm{cos}\:{x}}=\int\mathrm{sec}\:{x}\:{dx}= \\ $$$$=\int\mathrm{sec}\:{x}\:\frac{\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}}{\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}}{dx}\:\overset{\left[{t}=\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}\right]} {=} \\ $$$$=\int\frac{{dt}}{{t}}=\mathrm{ln}\:{t}\:= \\ $$$$=\mathrm{ln}\:\mid\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}\mid\:+{C} \\ $$
Answered by mr W last updated on 30/Sep/24
![=∫((d(sin x))/(1−sin^2 x)) =(1/2)[∫((d(sin x))/(1−sin x))+∫((d(sin x))/(1+sin x))] =(1/2)ln ((1+sin x)/(1−sin x))+C](https://www.tinkutara.com/question/Q212105.png)
$$=\int\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}−\mathrm{sin}\:{x}}+\int\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}+\mathrm{sin}\:{x}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}+{C} \\ $$