Question Number 211885 by Spillover last updated on 23/Sep/24
Answered by Frix last updated on 23/Sep/24
![∫(x^2 /((xsin x +cos x)^2 ))dx=(y/(xsin x +cos x))+C (d/dx)[(y/(xsin x +cos x))]=(x^2 /((xsin x +cos x)^2 )) ((y′(xsin x +cos x)−yxcos x)/((xsin x +cos x)^2 ))=(x^2 /((xsin x +cos x)^2 )) ⇒ y=sin x −xcos x ⇒ ∫(x^2 /((xsin x +cos x)^2 ))dx=((sin x −xcos x)/(xsin x +cos x))+C](https://www.tinkutara.com/question/Q211905.png)
$$\int\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{dx}=\frac{{y}}{{x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}+{C} \\ $$$$\frac{{d}}{{dx}}\left[\frac{{y}}{{x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}\right]=\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} } \\ $$$$\frac{{y}'\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)−{yx}\mathrm{cos}\:{x}}{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{y}=\mathrm{sin}\:{x}\:−{x}\mathrm{cos}\:{x} \\ $$$$\Rightarrow \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{dx}=\frac{\mathrm{sin}\:{x}\:−{x}\mathrm{cos}\:{x}}{{x}\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}+{C} \\ $$