Integration Questions

Question Number 205248 by universe last updated on 13/Mar/24

$$\:\:\:\int_{\mathrm{0}} ^{\pi} \:\frac{{x}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \left({x}\right)−{x}\mathrm{sin}\left({x}\right)−\mathrm{cos}\left({x}\right)−\mathrm{1}}{\left(\mathrm{1}+{x}\mathrm{sin}\left({x}\right)\right)^{\mathrm{2}} }{dx} \\$$

Answered by Berbere last updated on 13/Mar/24

$$\left(\frac{{f}\left({x}\right)}{\mathrm{1}+{xsin}\left({x}\right)}+{g}\left({x}\right)\right)^{'} =\frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)−{xsin}\left({x}\right)−{cos}\left({x}\right)−\mathrm{1}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} } \\$$$$\frac{{f}'\left(\mathrm{1}+{xsin}\left({x}\right)\right)−\left({sin}\left({x}\right)+{xcos}\left({x}\right)\right){f}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} }+{g}'\left({x}\right)=\frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)−{xsin}\left({x}\right)−{cos}\left({x}\right)−\mathrm{1}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} } \\$$$${f}=−{xcos}\left({x}\right) \\$$$$\Rightarrow\left(−{cos}\left({x}\right)+{xsin}\left({x}\right)\right)\left(\mathrm{1}+{xsin}\left({x}\right)\right)−\left({sin}\left({x}\right)+{xcos}\left({x}\right)\right)\left(−{xcos}\left({x}\right)\right. \\$$$${x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}\right)+{xsin}\left({x}\right)−{cos}\left({x}\right)={f}'\left(\mathrm{1}+{xsin}\left({x}\right)\right)−\left({sin}\left({x}\right)+{xcos}\left({x}\right)\right){f} \\$$$$\Rightarrow\frac{{f}'\left(\mathrm{1}+{xsin}\left({x}\right)\right)−\left({sin}\left({x}\right)+{xcos}\left({x}\right)\right){f}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} }+{g}' \\$$$$=\frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)−{cos}\left({x}\right)−{xsin}\left({x}\right)−\mathrm{1}+\mathrm{2}{xsin}\left({x}\right)+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}\right)+\mathrm{1}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} } \\$$$$=\frac{{d}}{{dx}}\left(\frac{−{xcos}\left({x}\right)}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)}\right)+\frac{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} } \\$$$$\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)−{xsin}\left({x}\right)−{cos}\left({x}\right)−\mathrm{1}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\pi} \left(\frac{−{xcos}\left({x}\right)}{\left(\mathrm{1}+{xsin}\left({x}\right)\right.}\overset{'} {\right)}−\mathrm{1}{dx} \\$$$$\\$$$$=\left[−\frac{{xcos}\left({x}\right)}{\mathrm{1}+{xsin}\left({x}\right)}−{x}\right]_{\mathrm{0}} ^{\pi} =\pi−\pi=\mathrm{0} \\$$$$\\$$$$\\$$$$\\$$

Commented by universe last updated on 14/Mar/24

$${thank}\:{u}\:{sir} \\$$