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
![(((f(x))/(1+xsin(x)))+g(x))^′ =((x^2 cos^2 (x)−xsin(x)−cos(x)−1)/((1+xsin(x))^2 )) ((f′(1+xsin(x))−(sin(x)+xcos(x))f)/((1+xsin(x))^2 ))+g′(x)=((x^2 cos^2 (x)−xsin(x)−cos(x)−1)/((1+xsin(x))^2 )) f=−xcos(x) ⇒(−cos(x)+xsin(x))(1+xsin(x))−(sin(x)+xcos(x))(−xcos(x) x^2 cos^2 (x)+x^2 sin^2 (x)+xsin(x)−cos(x)=f′(1+xsin(x))−(sin(x)+xcos(x))f ⇒((f′(1+xsin(x))−(sin(x)+xcos(x))f)/((1+xsin(x))^2 ))+g′ =((x^2 cos^2 (x)−cos(x)−xsin(x)−1+2xsin(x)+x^2 sin^2 (x)+1)/((1+xsin(x))^2 )) =(d/dx)(((−xcos(x))/((1+xsin(x)))))+(((1+xsin(x))^2 )/((1+xsin(x))^2 )) ∫_0 ^π ((x^2 cos^2 (x)−xsin(x)−cos(x)−1)/((1+xsin(x))^2 ))dx=∫_0 ^π (((−xcos(x))/((1+xsin(x))))^′ −1dx =[−((xcos(x))/(1+xsin(x)))−x]_0 ^π =π−π=0](https://www.tinkutara.com/question/Q205261.png)
$$\left(\frac{{f}\left({x}\right)}{\mathrm{1}+{xsin}\left({x}\right)}+{g}\left({x}\right)\right)^{'} =\frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)−{xsin}\left({x}\right)−{cos}\left({x}\right)−\mathrm{1}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$\frac{{f}'\left(\mathrm{1}+{xsin}\left({x}\right)\right)−\left({sin}\left({x}\right)+{xcos}\left({x}\right)\right){f}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} }+{g}'\left({x}\right)=\frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)−{xsin}\left({x}\right)−{cos}\left({x}\right)−\mathrm{1}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$${f}=−{xcos}\left({x}\right) \\ $$$$\Rightarrow\left(−{cos}\left({x}\right)+{xsin}\left({x}\right)\right)\left(\mathrm{1}+{xsin}\left({x}\right)\right)−\left({sin}\left({x}\right)+{xcos}\left({x}\right)\right)\left(−{xcos}\left({x}\right)\right. \\ $$$${x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}\right)+{xsin}\left({x}\right)−{cos}\left({x}\right)={f}'\left(\mathrm{1}+{xsin}\left({x}\right)\right)−\left({sin}\left({x}\right)+{xcos}\left({x}\right)\right){f} \\ $$$$\Rightarrow\frac{{f}'\left(\mathrm{1}+{xsin}\left({x}\right)\right)−\left({sin}\left({x}\right)+{xcos}\left({x}\right)\right){f}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} }+{g}' \\ $$$$=\frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)−{cos}\left({x}\right)−{xsin}\left({x}\right)−\mathrm{1}+\mathrm{2}{xsin}\left({x}\right)+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({x}\right)+\mathrm{1}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$=\frac{{d}}{{dx}}\left(\frac{−{xcos}\left({x}\right)}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)}\right)+\frac{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({x}\right)−{xsin}\left({x}\right)−{cos}\left({x}\right)−\mathrm{1}}{\left(\mathrm{1}+{xsin}\left({x}\right)\right)^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{\pi} \left(\frac{−{xcos}\left({x}\right)}{\left(\mathrm{1}+{xsin}\left({x}\right)\right.}\overset{'} {\right)}−\mathrm{1}{dx} \\ $$$$ \\ $$$$=\left[−\frac{{xcos}\left({x}\right)}{\mathrm{1}+{xsin}\left({x}\right)}−{x}\right]_{\mathrm{0}} ^{\pi} =\pi−\pi=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by universe last updated on 14/Mar/24
$${thank}\:{u}\:{sir} \\ $$