question 106075 again ∫((1+cosx)/((99cosx−70sinx+210)cosx−66sinx+110))dx=? using t=tan(x/2) I get −4∫(dt/(t^4 +8t^3 −22t^2 +272t−419)) can someone factorize the denominator?


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Question Number 106120 by Her_Majesty last updated on 02/Aug/20

$q u e s t i o n 106075 a g a i n$ $\int \frac{1 + c o s x}{(99 c o s x - 70 s i n x + 210) c o s x - 66 s i n x + 110} d x = ?$ $u s i n g t = t a n (x / 2) I g e t$ $- 4 \int \frac{d t}{t^{4} + 8 t^{3} - 22 t^{2} + 272 t - 419}$ $c a n s o m e o n e f a c t o r i z e t h e d e n o m i n a t o r ?$
Commented by Sarah85 last updated on 03/Aug/20
$typo: t^4 −8t^3 −22t^2 +272t−419 in this case t= { ((2−(√3)−(√5)−(√(15)))),((2+(√3)+(√5)−(√(15)))),((2+(√3)−(√5)+(√(15)))),((2−(√3)+(√5)+(√(15)))) :}$
$typo : t^{4} - 8 t^{3} - 22 t^{2} + 272 t - 419$ $in this case$ $t = {\begin{cases} 2 - \sqrt{3} - \sqrt{5} - \sqrt{15} \\ 2 + \sqrt{3} + \sqrt{5} - \sqrt{15} \\ 2 + \sqrt{3} - \sqrt{5} + \sqrt{15} \\ 2 - \sqrt{3} + \sqrt{5} + \sqrt{15} \end{cases}$
Commented by Her_Majesty last updated on 03/Aug/20

$t h a n k s b u t h o w y o u g o t i t ?$
Commented by 1549442205PVT last updated on 03/Aug/20

$t^{4} - {8 t}^{3} - {22 t}^{2} + 272 t - 419 = 0$ $\Leftrightarrow [t^{2} - (4 - 2 \sqrt{15}) t + 11 - 6 \sqrt{15}] \times$ $[t^{2} - (4 + 2 \sqrt{15}) t + 11 + 6 \sqrt{15}] = 0$ $\Leftrightarrow [t^{2} - (4 - 2 \sqrt{5}) t - (9 + 10 \sqrt{5})] \times$ $[t^{2} - (4 + 2 \sqrt{5}) t - (9 - 10 \sqrt{5})] = 0$
	Answered by 1549442205PVT last updated on 03/Aug/20

	$t^{4} + 8 t^{3} - 22 t^{2} + 272 t - 419$ $= (t^{2} + 10 . 374065851 t - 19.19728528) \times$ $(t^{2} - 2.374065851 + 21.82600076)$ $two real roots be$ $x_{1} = 1.602856312, x_{2} = - 11.97692216$
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