cos-4x-cos-2x-sin-4x-cos-2x-dx-

Question Number 82954 by john santu last updated on 26/Feb/20

\int \frac{\cos 4 x - \cos 2 x}{\sin 4 x - \cos 2 x} dx

Commented by john santu last updated on 26/Feb/20

let u = \cos 2 x \Rightarrow dx = - \frac{du}{2 \sqrt{1 - u^{2}}}

\Rightarrow \int \frac{{2 u}^{2} - u - 1}{2 u \sqrt{1 - u^{2}} - u} \times \frac{du}{- 2 \sqrt{1 - u^{2}}}

\Rightarrow \int \frac{{2 u}^{2} - u - 1}{- 4 u (1 - u^{2}) + 2 u \sqrt{1 - u^{2}}} du

stuck

Commented by MJS last updated on 26/Feb/20

always try t = \tan \frac{x}{2} \Leftrightarrow x = 2 \arctan t but in

this case we have only 2 n x {it}^{'} s enough to

let t = \tan x

Weierstrass substitution :

t = \tan \frac{x}{2} \to d x = \frac{2 d t}{1 + t^{2}}; \sin x = \frac{2 t}{1 + t^{2}}; \cos x = \frac{1 - t^{2}}{1 + t^{2}}

Commented by john santu last updated on 26/Feb/20

o yes sir . i^{'} m forgot this method .

thank you mister mjs

Answered by MJS last updated on 26/Feb/20

\int \frac{\cos 4 x - \cos 2 x}{\sin 4 x - \cos 2 x} d x =

[t = \tan x \to d x = \frac{d t}{t^{2} + 1}]

= 2 \int \frac{(t^{2} - 3) t^{2}}{(t - 1) (t + 1) (t - 2 - \sqrt{3}) (t - 2 + \sqrt{3}) (t^{2} + 1)} d t =

= \frac{1}{2} \int \frac{d t}{t - 1} + \frac{1}{6} \int \frac{d t}{t + 1} + \frac{1 + \sqrt{3}}{6} \int \frac{d t}{t - 2 - \sqrt{3}} + \frac{1 - \sqrt{3}}{6} \int \frac{d t}{t - 2 + \sqrt{3}} - \int \frac{t}{t^{2} + 1} d t

and all of them are easy to solve

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