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Question Number 211798 by Spillover last updated on 21/Sep/24

	Answered by Ghisom last updated on 22/Sep/24

	$\frac{e^{\cos x} (x \sin^{3} x + \cos x)}{\sin^{2} x} =$ $= e^{\cos x} x \sin x + \frac{e^{\cos x} \cos x}{\sin^{2} x} =$ $= (e^{\cos x} x \sin x) - e^{\cos x} + \frac{e^{\cos x} \cos x}{\sin^{2} x} + e^{\cos x} =$ $= e^{\cos x} (x \sin x - 1) + e^{\cos x} (\frac{\cos x}{\sin^{2} x} + 1)$ $1 .$ $\frac{d}{d x} [e^{\cos x} y_{1}] = e^{\cos x} (x \sin x - 1)$ $e^{\cos x} (y_{1}^{'} - y_{1} \sin x) = e^{\cos x} (x \sin x - 1)$ $\Rightarrow y_{1} = - x$ $\int e^{\cos x} (x \sin x - 1) d x = - e^{\cos x} x ★$ $2 .$ $\int e^{\cos x} (\frac{\cos x}{\sin^{2} x} + 1) d x =$ $= \int e^{\cos x} \frac{\cos x}{\sin^{2} x} d x + \int e^{\cos x} d x =$ $u^{'} = \frac{\cos x}{\sin^{2} x} \to u = - \frac{1}{\sin x}$ $v = e^{\cos x} \to v^{'} = - e^{\cos x} \sin x$ $= - \frac{e^{\cos x}}{\sin x} - \int e^{\cos x} d x + \int e^{\cos x} d x =$ $= - \frac{e^{\cos x}}{\sin x} ★$ $=====================$ $\int \frac{e^{\cos x} (x \sin^{3} x + \cos x)}{\sin^{2} x} d x =$ $= - e^{\cos x} (x + \frac{1}{\sin x}) + C$
	Commented by Ghisom last updated on 22/Sep/24

	$\frac{d}{d x} [- e^{\cos x} (x + \frac{1}{\sin x})] = . . .$ $. . . = \frac{e^{\cos x} (x \sin^{3} x + \cos x)}{\sin^{2} x}$ $so {what}^{'} s wrong in your opinion ?$
	Commented by TonyCWX08 last updated on 22/Sep/24

	$S o m e t h i n g i s n o t q u i t e r i g h t h e r e . . .$ $C h e c k a g a i n ?$
	Commented by TonyCWX08 last updated on 22/Sep/24

	$O k a y$
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