∫ ((2xsin 2x)/((2x−sin 2x)^2 )) dx ?


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Question Number 78581 by john santu last updated on 18/Jan/20

$\int \frac{2 x \sin 2 x}{{(2 x - \sin 2 x)}^{2}} d x ?$
Commented by john santu last updated on 19/Jan/20

$c o n s i d e r$ $\frac{2 x \sin 2 x}{{(2 x - \sin 2 x)}^{2}} \times \frac{1 - \cos 2 x}{1 - \cos 2 x} =$ $\frac{2 x \sin 2 x}{1 - \cos 2 x} \times \frac{1 - \cos 2 x}{{(2 x - \sin 2 x)}^{2}}$ $n o w u s e i n t e g r a t i o n b y p a r t s$ $w h e r e u = \frac{2 x \sin 2 x}{1 - \cos 2 x} \Rightarrow d u = \frac{\sin 2 x - 2 x}{1 - \cos 2 x}$ $d v = \frac{1 - \cos 2 x}{(2 x - \sin 2 x)} d x \Rightarrow v = \frac{1}{\sin 2 x - 2 x}$ $I = u . v - \int v d u$ $I = \frac{2 x \sin 2 x}{1 - \cos 2 x} . \frac{1}{\sin 2 x - 2 x} - \int (\frac{1}{\sin 2 x - 2 x}) . \frac{\sin 2 x - 2 x}{1 - \cos 2 x} d x$ $I = \frac{2 x \sin 2 x}{(1 - \cos 2 x) (\sin 2 x - 2 x)} - \int \frac{d x}{1 - \cos 2 x}$ $I = \frac{2 x \sin 2 x}{(1 - \cos 2 x) (\sin 2 x - 2 x)} - \int \frac{d x}{2 \sin^{2} x}$ $I = \frac{2 x \sin 2 x}{(1 - \cos 2 x) (\sin 2 x - 2 x)} + \frac{\cot x}{2} + c$
Commented by john santu last updated on 19/Jan/20

$I = \frac{4 x \sin x \cos x}{(2 \sin^{2} x) (\sin 2 x - 2 x)} + \frac{\cos x}{2 \sin x}$ $I = \frac{4 x \cos x}{2 \sin x (\sin 2 x - 2 x)} + \frac{\cos x}{2 \sin x}$ $I = \frac{4 x \cos x + \sin 2 x \cos x - 2 x \cos x}{2 \sin x (\sin 2 x - 2 x)}$ $I = \frac{2 x \cos x + \sin 2 x \cos x}{2 \sin x (\sin 2 x - 2 x)} + c$ $I = \frac{\cos x (2 x + \sin 2 x)}{2 \sin x (\sin 2 x - 2 x)} + c$
Commented by jagoll last updated on 19/Jan/20

$waww$
Commented by john santu last updated on 19/Jan/20

$t h a n k s$
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