x-dx-cot-x-tan-x-2-

Question Number 131552 by liberty last updated on 06/Feb/21

\int \frac{x dx}{{(\cot x + \tan x)}^{2}} ?

Answered by EDWIN88 last updated on 06/Feb/21

\Leftrightarrow \cot x + \tan x = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2 x}

\Leftrightarrow \frac{1}{{(\cot x + \tan x)}^{2}} = \frac{\sin^{2} 2 x}{4}

Now E = \int \frac{x dx}{{(\cot x + \tan x)}^{2}} = \frac{1}{4} \int xsin^{2} 2 xdx

E = \frac{1}{8} \int (x - xcos 4 x) dx = \frac{1}{16} x^{2} - \frac{1}{8} \int x \cos 4 x dx

E = \frac{1}{16} x^{2} - \frac{1}{8} [\frac{1}{4} x \sin 4 x - \frac{1}{4} \int \sin 4 x dx]

E = \frac{1}{16} x^{2} - \frac{1}{32} x \sin 4 x - \frac{1}{128} \cos 4 x + c

E = \frac{{8 x}^{2} - 4 x \sin 4 x - \cos 4 x}{128} + c

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