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