Question Number 211798 by Spillover last updated on 21/Sep/24
Answered by Ghisom last updated on 22/Sep/24
![((e^(cos x) (xsin^3 x +cos x))/(sin^2 x))= =e^(cos x) xsin x +((e^(cos x) cos x)/(sin^2 x))= =(e^(cos x) xsin x)−e^(cos x) +((e^(cos x) cos x)/(sin^2 x))+e^(cos x) = =e^(cos x) (xsin x −1)+e^(cos x) (((cos x)/(sin^2 x))+1) 1. (d/dx)[e^(cos x) y_1 ]=e^(cos x) (xsin x −1) e^(cos x) (y_1 ′−y_1 sin x)=e^(cos x) (xsin x −1) ⇒ y_1 =−x ∫e^(cos x) (xsin x −1)dx=−e^(cos x) x ★ 2. ∫e^(cos x) (((cos x)/(sin^2 x))+1)dx= =∫e^(cos x) ((cos x)/(sin^2 x))dx+∫e^(cos x) dx= u′=((cos x)/(sin^2 x)) → u=−(1/(sin x)) v=e^(cos x) → v′=−e^(cos x) sin x =−(e^(cos x) /(sin x))−∫e^(cos x) dx+∫e^(cos x) dx= =−(e^(cos x) /(sin x)) ★ ===================== ∫((e^(cos x) (xsin^3 x +cos x))/(sin^2 x))dx= =−e^(cos x) (x+(1/(sin x)))+C](https://www.tinkutara.com/question/Q211816.png)
$$\frac{\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}\mathrm{sin}^{\mathrm{3}} \:{x}\:+\mathrm{cos}\:{x}\right)}{\mathrm{sin}^{\mathrm{2}} \:{x}}= \\ $$$$=\mathrm{e}^{\mathrm{cos}\:{x}} {x}\mathrm{sin}\:{x}\:+\frac{\mathrm{e}^{\mathrm{cos}\:{x}} \mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}= \\ $$$$=\left(\mathrm{e}^{\mathrm{cos}\:{x}} {x}\mathrm{sin}\:{x}\right)−\mathrm{e}^{\mathrm{cos}\:{x}} +\frac{\mathrm{e}^{\mathrm{cos}\:{x}} \mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}+\mathrm{e}^{\mathrm{cos}\:{x}} = \\ $$$$=\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}\mathrm{sin}\:{x}\:−\mathrm{1}\right)+\mathrm{e}^{\mathrm{cos}\:{x}} \left(\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}+\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{1}. \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{e}^{\mathrm{cos}\:{x}} {y}_{\mathrm{1}} \right]=\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}\mathrm{sin}\:{x}\:−\mathrm{1}\right) \\ $$$$\mathrm{e}^{\mathrm{cos}\:{x}} \left({y}_{\mathrm{1}} '−{y}_{\mathrm{1}} \mathrm{sin}\:{x}\right)=\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}\mathrm{sin}\:{x}\:−\mathrm{1}\right) \\ $$$$\Rightarrow\:{y}_{\mathrm{1}} =−{x} \\ $$$$\int\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}\mathrm{sin}\:{x}\:−\mathrm{1}\right){dx}=−\mathrm{e}^{\mathrm{cos}\:{x}} {x}\:\bigstar \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\int\mathrm{e}^{\mathrm{cos}\:{x}} \left(\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}+\mathrm{1}\right){dx}= \\ $$$$=\int\mathrm{e}^{\mathrm{cos}\:{x}} \frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}+\int\mathrm{e}^{\mathrm{cos}\:{x}} {dx}= \\ $$$$\:\:\:\:\:{u}'=\frac{\mathrm{cos}\:{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}\:\rightarrow\:{u}=−\frac{\mathrm{1}}{\mathrm{sin}\:{x}} \\ $$$$\:\:\:\:\:{v}=\mathrm{e}^{\mathrm{cos}\:{x}} \:\rightarrow\:{v}'=−\mathrm{e}^{\mathrm{cos}\:{x}} \mathrm{sin}\:{x} \\ $$$$=−\frac{\mathrm{e}^{\mathrm{cos}\:{x}} }{\mathrm{sin}\:{x}}−\int\mathrm{e}^{\mathrm{cos}\:{x}} {dx}+\int\mathrm{e}^{\mathrm{cos}\:{x}} {dx}= \\ $$$$=−\frac{\mathrm{e}^{\mathrm{cos}\:{x}} }{\mathrm{sin}\:{x}}\:\bigstar \\ $$$$===================== \\ $$$$\int\frac{\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}\mathrm{sin}^{\mathrm{3}} \:{x}\:+\mathrm{cos}\:{x}\right)}{\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$=−\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}+\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\right)+{C} \\ $$
Commented by Ghisom last updated on 22/Sep/24
![(d/dx)[−e^(cos x) (x+(1/(sin x)))]=... ...=((e^(cos x) (xsin^3 x +cos x))/(sin^2 x)) so what′s wrong in your opinion?](https://www.tinkutara.com/question/Q211829.png)
$$\frac{{d}}{{dx}}\left[−\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}+\frac{\mathrm{1}}{\mathrm{sin}\:{x}}\right)\right]=… \\ $$$$…=\frac{\mathrm{e}^{\mathrm{cos}\:{x}} \left({x}\mathrm{sin}^{\mathrm{3}} \:{x}\:+\mathrm{cos}\:{x}\right)}{\mathrm{sin}^{\mathrm{2}} \:{x}} \\ $$$$\mathrm{so}\:\mathrm{what}'\mathrm{s}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{your}\:\mathrm{opinion}? \\ $$
Commented by TonyCWX08 last updated on 22/Sep/24
$${Something}\:{is}\:{not}\:{quite}\:{right}\:{here}… \\ $$$${Check}\:{again}? \\ $$
Commented by TonyCWX08 last updated on 22/Sep/24
$${Okay} \\ $$