Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 212099 by vahid last updated on 30/Sep/24

Answered by mehdee7396 last updated on 30/Sep/24

∫((1+tan^2 (x/2))/(1−tan^2 (x/2)))dx ; let tan(x/2)=u =∫((2u)/(1−u^2 ))du=ln((1+u)/(1−u))+c =ln(tan((π/4)+(x/2)))+c ✓

$$\int\frac{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}\:\:\:\:\:;\:{let}\:\:\:{tan}\frac{{x}}{\mathrm{2}}={u} \\ $$$$=\int\frac{\mathrm{2}{u}}{\mathrm{1}−{u}^{\mathrm{2}} }{du}={ln}\frac{\mathrm{1}+{u}}{\mathrm{1}−{u}}+{c} \\ $$$$={ln}\left({tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)\right)+{c}\:\:\checkmark \\ $$$$ \\ $$$$ \\ $$

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

$$\int\frac{{dx}}{\mathrm{cos}\:{x}}=\int\mathrm{sec}\:{x}\:{dx}= \\ $$$$=\int\mathrm{sec}\:{x}\:\frac{\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}}{\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}}{dx}\:\overset{\left[{t}=\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}\right]} {=} \\ $$$$=\int\frac{{dt}}{{t}}=\mathrm{ln}\:{t}\:= \\ $$$$=\mathrm{ln}\:\mid\mathrm{tan}\:{x}\:+\mathrm{sec}\:{x}\mid\:+{C} \\ $$

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

$$=\int\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}−\mathrm{sin}\:{x}}+\int\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}+\mathrm{sin}\:{x}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}+{C} \\ $$

Commented by MathematicalUser2357 last updated on 02/Oct/24

1/2 ln|(1+sinx)/(1-sinx)|+C*

1/2 ln|(1+sinx)/(1-sinx)|+C*

Commented by Frix last updated on 02/Oct/24

What? Why? −1≤sin x ≤1 1. 0≤1+sin x ≤2 2. −2≤−1+sin x≤0 ⇔ 0≤1−sin x ≤2 ⇒ ∣((1+sin x)/(1−sin x))∣=((1+sin x)/(1−sin x))

$$\mathrm{What}?\:\mathrm{Why}? \\ $$$$−\mathrm{1}\leqslant\mathrm{sin}\:{x}\:\leqslant\mathrm{1} \\ $$$$\mathrm{1}.\:\:\:\:\:\:\mathrm{0}\leqslant\mathrm{1}+\mathrm{sin}\:{x}\:\leqslant\mathrm{2} \\ $$$$\mathrm{2}.\:\:\:\:\:−\mathrm{2}\leqslant−\mathrm{1}+\mathrm{sin}\:{x}\leqslant\mathrm{0}\:\Leftrightarrow\:\mathrm{0}\leqslant\mathrm{1}−\mathrm{sin}\:{x}\:\leqslant\mathrm{2} \\ $$$$\Rightarrow \\ $$$$\mid\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\mid=\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com