Integration Questions

Question Number 144876 by qaz last updated on 30/Jun/21

$$\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}−\frac{\mathrm{1}}{\mathrm{x}}\right)\mathrm{dx}=\mathrm{ln}\left(\mathrm{2tan}\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\$$

Answered by mindispower last updated on 30/Jun/21

$$\\$$$${sin}\left({x}\right)=\frac{\mathrm{2}{tg}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+{tg}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)} \\$$$$\Leftrightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}+{tg}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{tg}\left(\frac{{x}}{\mathrm{2}}\right)}−\frac{\mathrm{1}}{{x}}{dx} \\$$$$=\int_{{t}} ^{\mathrm{1}} \frac{\mathrm{1}}{{tg}\left(\frac{{x}}{\mathrm{2}}\right)}.{d}\left({tg}\left(\frac{{x}}{\mathrm{2}}\right)\right)−\int_{{t}} ^{\mathrm{1}} \frac{{dx}}{{x}}={f}\left({t}\right) \\$$$$\left.={ln}\left({tg}\left(\frac{{x}}{\mathrm{2}}\right)\right)−{ln}\left({x}\right)\right]_{{t}} ^{\mathrm{1}} \\$$$$={ln}\left({tg}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+{ln}\left(\frac{{t}}{{tg}\left(\frac{{t}}{\mathrm{2}}\right)}\right) \\$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{sin}\left({x}\right)}−\frac{\mathrm{1}}{{x}}{dx} \\$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right)={ln}\left({tg}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+{ln}\left(\mathrm{2}\right) \\$$$$={ln}\left(\mathrm{2}{tg}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\$$$$\\$$

Commented by mnjuly1970 last updated on 30/Jun/21

$$\:\:\:\:{very}\:{nice}\:{solution}... \\$$

Commented by qaz last updated on 30/Jun/21

$$\mathrm{thanks}\:\mathrm{a}\:\mathrm{lot} \\$$

Commented by mindispower last updated on 04/Jul/21

$${pleasur}\:{sir} \\$$