Question Number 138703 by mathocean1 last updated on 16/Apr/21 | ||
$${Calculate} \\ $$$$\underset{−\frac{\pi}{\mathrm{6}}} {\overset{\mathrm{0}} {\int}}\frac{{cos}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{2}{sinx}}{dx} \\ $$ | ||
Commented by SanyamJoshi last updated on 17/Apr/21 | ||
$${Calculate} \\ $$$$\underset{−\frac{\pi}{\mathrm{6}}} {\overset{\mathrm{0}} {\int}}\frac{{cos}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{2}{sinx}}{dx} \\ $$ | ||
Answered by Mathspace last updated on 16/Apr/21 | ||
$${I}=\int_{−\frac{\pi}{\mathrm{6}}} ^{\mathrm{0}} \:\frac{{cos}^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{2}{sinx}}{dx} \\ $$$$=_{{x}=−{t}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{cos}^{\mathrm{2}} {t}}{\mathrm{1}+\mathrm{2}{sint}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{\mathrm{1}−{sin}^{\mathrm{2}} {t}}{\mathrm{1}+\mathrm{2}{sint}}{dt} \\ $$$${we}\:{have}\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{4}}\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{2}{x}+\mathrm{1}}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{x}−\mathrm{1}\right)+\frac{\mathrm{3}}{\mathrm{4}\left(\mathrm{2}{x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \left(\mathrm{2}{sint}−\mathrm{1}\right){dt} \\ $$$$+\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{dt}}{\mathrm{2}{sint}\:+\mathrm{1}}\left(={J}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[−{cost}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} +\frac{\mathrm{1}}{\mathrm{4}}.\frac{\pi}{\mathrm{6}} \\ $$$$+\frac{\mathrm{3}}{\mathrm{4}}{J} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{24}}+\frac{\mathrm{3}}{\mathrm{4}}{J} \\ $$$${J}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\frac{{dt}}{\mathrm{2}{sint}\:+\mathrm{1}} \\ $$$$=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={y}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \:\:\:\frac{\mathrm{2}{dy}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{2}\frac{\mathrm{2}{y}}{\mathrm{1}+{y}^{\mathrm{2}} }+\mathrm{1}\right)} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \:\:\:\:\frac{{dy}}{\mathrm{4}{y}+{y}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \:\:\:\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{1}} \\ $$$$\Delta^{'} =\mathrm{2}^{\mathrm{2}} −\mathrm{1}=\mathrm{3}\:\Rightarrow{y}_{\mathrm{1}} =−\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${y}_{\mathrm{2}} =−\mathrm{2}−\sqrt{\mathrm{3}}\:\Rightarrow \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \:\frac{{dy}}{\left({y}−{y}_{\mathrm{1}} \right)\left({y}−{y}_{\mathrm{2}} \right)} \\ $$$$=\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \:\:\left(\frac{\mathrm{1}}{{y}−{y}_{\mathrm{1}} }−\frac{\mathrm{1}}{{y}−{y}_{\mathrm{2}} }\right){dy} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left[{log}\mid\frac{{y}+\mathrm{2}−\sqrt{\mathrm{3}}}{{y}+\mathrm{2}+\sqrt{\mathrm{3}}}\mid\right]_{\mathrm{0}} ^{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{{log}\left(\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}\right)−{log}\left(\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right)\right\} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}+\frac{\pi}{\mathrm{24}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left\{{log}\left(\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−{log}\left(\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{2}+\sqrt{\mathrm{3}}}\right)\right\} \\ $$ | ||