x-dx-cot-x-tan-x-2-a-x-16-x-sin-4x-32-cos-4x-128-c-b-x-16-x-sin-4x-32-cos-4x-128-c-c-x-16-xsin-4x-64-cos-4x-128-c-d-x-16-xcos-

Question Number 103512 by bemath last updated on 15/Jul/20

\int \frac{x d x}{{(\cot x + \tan x)}^{2}} =

(a) \frac{x}{16} - \frac{x \sin 4 x}{32} - \frac{\cos 4 x}{128} + c

(b) \frac{x}{16} + \frac{x \sin 4 x}{32} - \frac{\cos 4 x}{128} + c

(c) \frac{x}{16} + \frac{x \sin 4 x}{64} + \frac{\cos 4 x}{128} + c

(d) \frac{x}{16} + \frac{x \cos 4 x}{32} + \frac{\sin 4 x}{128} + c

Commented by bobhans last updated on 15/Jul/20

n o t h i n g a n s w e r t o o

Commented by bemath last updated on 15/Jul/20

y e s . i t h i n k t h e q u e s t i o n w r o n g

Answered by Dwaipayan Shikari last updated on 15/Jul/20

\int \frac{x}{{(\frac{c o s x}{s i n x} + \frac{s i n x}{c o s x})}^{2}} d x = \frac{1}{4} \int {x s i n}^{2} 2 x = \int \frac{x}{8} (1 - c o s 4 x)

\int \frac{x}{8} - \frac{1}{8} \int x c o s 4 x

\frac{x^{2}}{16} - \frac{1}{32} x s i n 4 x + \frac{1}{32} \int s i n 4 x

\frac{x^{2}}{16} - \frac{1}{32} x s i n 4 x - \frac{1}{128} c o s 4 x + Constant

Answered by bramlex last updated on 15/Jul/20

\cot x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}

= \frac{2}{\sin 2 x}; I = \int \frac{x dx}{(\frac{4}{\sin^{2} 2 x})}

= \frac{1}{4} \int x (\frac{1}{2} - \frac{1}{2} \cos 4 x) dx

= \frac{1}{8} \int (x - xcos 4 x) dx

= \frac{1}{8} (\frac{x^{2}}{2} - \frac{xsin 4 x}{4} - \frac{\cos 4 x}{16}) + c

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

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