∫_0 ^π ((cos(x))/(1+2sin(2x)))dx


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 34992 by MJS last updated on 14/May/18

$\underset{0}{\int^{π}} \frac{\cos (x)}{1 + 2 \sin (2 x)} d x$
Commented by math khazana by abdo last updated on 14/May/18
$let put I = ∫_0 ^π ((cosx)/(1+2sin(2x)))dx I = ∫_0 ^π ((cosx)/(1+4cosx sinx))dx .changement tan((x/2))=t give I = ∫_0 ^∞ (((1−t^2 )/(1+t^2 ))/(1+4 ((1−t^2 )/(1+t^2 )) ((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = 2∫_0 ^∞ ((1−t^2 )/((1+t^2 )^2 ( 1+((8t(1−t^2 ))/((1+t^2 )^2 )))))dt = 2 ∫_0 ^∞ ((1−t^2 )/((1+t^2 )^2 +8t(1−t^2 )))dt = 2 ∫_0 ^∞ ((1−t^2 )/(t^4 +2t^2 +1 +8t −8t^3 ))dt = 2 ∫_0 ^∞ ((1−t^2 )/(t^4 −8t^3 +2t^2 +8t +1))dt let drcompose F(t) = ((1−t^2 )/(t^4 −8t^3 +2t^2 +8t +1)) the roots of p(t)=t^4 −8t^3 +2t^2 +8t +1 are t_1 ∼7,59 t_2 ∼1,3 t_3 ∼−0,76 t_4 ∼−0,13 ⇒ F(t) = ((1−t^2 )/((t −t_1 )(t −t_2 )(t −t_3 )(t −t_4 ))) = (a/(t−t_1 )) +(b/(t−t_2 )) +(c/(t−t_3 )) +(d/(t −t_4 )) a=lim_(t→t_1 ) (t−t_1 )F(t)= ((1−t_1 ^2 )/((t_1 −t_2 )(t_1 −t_3 )(t_1 −t_4 ))) =ξ(t_1 ) (1−t_1 ^2 ) with ξ(t_1 )= (1/((t_1 −t_2 )(t_1 −t_3 )(t_1 −t_4 ))) b =ξ(t_2 )(1−t_2 ^2 ) with ξ(t_2 )= (1/((t_2 −t_1 )(t_2 −t_3 )(t_2 −t_4 ))) c =ξ(t_3 )(1−t_3 ^2 ) d=ξ(t_4 )(1−t_4 ^2 ) ⇒ I =2 ∫_0 ^∞ F(t)dt = 2 [ aln∣t−t_1 ∣ +b ln∣t−t_2 ∣ +c ln∣t−t_3 ) +d ln∣t−t_4 ∣]_0 ^(+∞) =2[ ln∣(((t−t_1 )a)/((t −t_2 )^b )) ∣ + ln∣ (((t−t_3 )^c )/((t−t_4 )^d ))∣]_0 ^(+∞) =2{−ln∣(((−t_1 )^a )/((−t_2 )^b )) ∣−ln∣(((−t_3 )^c )/((−t_4 )^d ))∣} =2{ bln∣t_2 ∣ −aln∣t_1 ∣ +dln∣t_4 ∣ −c ln∣t_3 ∣}... the approximate value of I isdetermined...$
$l e t p u t I = \int_{0}^{π} \frac{c o s x}{1 + 2 s i n (2 x)} d x$ $I = \int_{0}^{π} \frac{c o s x}{1 + 4 c o s x s i n x} d x . c h a n g e m e n t t a n (\frac{x}{2}) = t$ $g i v e I = \int_{0}^{\infty} \frac{\frac{1 - t^{2}}{1 + t^{2}}}{1 + 4 \frac{1 - t^{2}}{1 + t^{2}} \frac{2 t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}$ $= 2 \int_{0}^{\infty} \frac{1 - t^{2}}{{(1 + t^{2})}^{2} (1 + \frac{8 t (1 - t^{2})}{{(1 + t^{2})}^{2}})} d t$ $= 2 \int_{0}^{\infty} \frac{1 - t^{2}}{{(1 + t^{2})}^{2} + 8 t (1 - t^{2})} d t$ $= 2 \int_{0}^{\infty} \frac{1 - t^{2}}{t^{4} + 2 t^{2} + 1 + 8 t - 8 t^{3}} d t$ $= 2 \int_{0}^{\infty} \frac{1 - t^{2}}{t^{4} - 8 t^{3} + 2 t^{2} + 8 t + 1} d t l e t d r c o m p o s e$ $F (t) = \frac{1 - t^{2}}{t^{4} - 8 t^{3} + 2 t^{2} + 8 t + 1}$ $t h e r o o t s o f p (t) = t^{4} - 8 t^{3} + 2 t^{2} + 8 t + 1 a r e$ $t_{1} \sim 7, 59 t_{2} \sim 1, 3 t_{3} \sim - 0, 76 t_{4} \sim - 0, 13 \Rightarrow$ $F (t) = \frac{1 - t^{2}}{(t - t_{1}) (t - t_{2}) (t - t_{3}) (t - t_{4})}$ $= \frac{a}{t - t_{1}} + \frac{b}{t - t_{2}} + \frac{c}{t - t_{3}} + \frac{d}{t - t_{4}}$ $a = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{1 - t_{1}^{2}}{(t_{1} - t_{2}) (t_{1} - t_{3}) (t_{1} - t_{4})}$ $= ξ (t_{1}) (1 - t_{1}^{2}) w i t h ξ (t_{1}) = \frac{1}{(t_{1} - t_{2}) (t_{1} - t_{3}) (t_{1} - t_{4})}$ $b = ξ (t_{2}) (1 - t_{2}^{2}) w i t h ξ (t_{2}) = \frac{1}{(t_{2} - t_{1}) (t_{2} - t_{3}) (t_{2} - t_{4})}$ $c = ξ (t_{3}) (1 - t_{3}^{2})$ $d = ξ (t_{4}) (1 - t_{4}^{2}) \Rightarrow$ $I = 2 \int_{0}^{\infty} F (t) d t$ ${= 2 [a l n ∣ t - t_{1} ∣ + b l n ∣ t - t_{2} ∣ + c l n ∣ t - t_{3}) + d l n ∣ t - t_{4} ∣]}_{0}^{+ \infty}$ $= 2 {[l n ∣ \frac{(t - t_{1}) a}{{(t - t_{2})}^{b}} ∣ + l n ∣ \frac{{(t - t_{3})}^{c}}{{(t - t_{4})}^{d}} ∣]}_{0}^{+ \infty}$ $= 2 {- l n ∣ \frac{{(- t_{1})}^{a}}{{(- t_{2})}^{b}} ∣ - l n ∣ \frac{{(- t_{3})}^{c}}{{(- t_{4})}^{d}} ∣}$ $= 2 {b l n ∣ t_{2} ∣ - a l n ∣ t_{1} ∣ + d l n ∣ t_{4} ∣ - c l n ∣ t_{3} ∣} . . .$ $t h e a p p r o x i m a t e v a l u e o f I i s d e t e r m i n e d . . .$
	Answered by MJS last updated on 14/May/18

	$\underset{0}{\int^{π}} \frac{\cos (x)}{1 + 2 \sin (2 x)} d x = \underset{- \frac{π}{2}}{\int^{\frac{π}{2}}} \frac{\sin (x)}{2 \sin (2 x) - 1} d x =$ $= \underset{- \frac{π}{2}}{\int^{\frac{π}{2}}} \frac{\sin (x)}{4 \sin (x) \cos (x) - 1} d x =$ $[\begin{matrix} t = \tan (\frac{x}{2}) \to d x = \frac{2 d t}{t^{2} + 1} \\ \sin (x) = \frac{2 t}{t^{2} + 1}; \cos (x) = - \frac{t^{2} - 1}{t^{2} + 1} \end{matrix}]$ $= - 4 \underset{- 1}{\int^{1}} \frac{t}{t^{4} + 8 t^{3} + 2 t^{2} - 8 t + 1} d t =$ $[\begin{matrix} t^{4} + 8 t^{3} + 2 t^{2} - 8 t + 1 = 0 \\ t_{1} = - 2 - \sqrt{6} - \sqrt{3} - \sqrt{2} \\ t_{2} = - 2 - \sqrt{6} + \sqrt{3} + \sqrt{2} \\ t_{3} = - 2 + \sqrt{6} - \sqrt{3} + \sqrt{2} \\ t_{4} = - 2 + \sqrt{6} + \sqrt{3} - \sqrt{2} \end{matrix}]$ $= - 4 \underset{- 1}{\int^{1}} (\frac{A}{t - t_{1}} + \frac{B}{t - t_{2}} + \frac{C}{t - t_{3}} + \frac{D}{t - t_{4}}) d t =$ $= - 4 {[A \ln ∣ t - t_{1} ∣ + B \ln ∣ t - t_{2} ∣ + C \ln ∣ t - t_{3} ∣ + D \ln ∣ t - t_{4} ∣]}_{- 1}^{1} =$ $Missing \left or extra \right$ $[\begin{matrix} A = \frac{t_{1}}{(t_{1} - t_{2}) (t_{1} - t_{3}) (t_{1} - t_{4})} = \frac{- \sqrt{6} + 3 \sqrt{2}}{96} \\ B = \frac{t_{2}}{(t_{2} - t_{1}) (t_{2} - t_{3}) (t_{2} - t_{4})} = \frac{- \sqrt{6} - 3 \sqrt{2}}{96} \\ C = \frac{t_{3}}{(t_{3} - t_{1}) (t_{3} - t_{2}) (t_{3} - t_{4})} = \frac{\sqrt{6} - 3 \sqrt{2}}{96} \\ D = \frac{t_{4}}{(t_{4} - t_{1}) (t_{4} - t_{2}) (t_{4} - t_{3})} = \frac{\sqrt{6} + 3 \sqrt{2}}{96} \end{matrix}]$ $Missing \left or extra \right$ $\approx 1.091166$
	Commented by MJS last updated on 14/May/18

	$I {couldn}^{'} t get this out of my head . . .$ $this should be the solution, please check for$ $mistakes of method .$ $I can guarantee the correctness of the zeros$ $t_{1}, t_{2}, t_{3}, t_{4} and the coefficients A, B, C, D .$
	Commented by ajfour last updated on 14/May/18

	$f o r x = \frac{π}{12}, \frac{\sin x}{2 \sin 2 x - 1} i s n o t d e f i n e d .$
	Commented by MJS last updated on 14/May/18

	$but {it}^{'} s the Cauchy principal value :$ $\int \frac{p}{x - q} d x = p \ln ∣ x - q ∣$ $a < q < b; h > 0$ $\underset{a}{\int^{b}} \frac{p}{x - q} d x = \lim_{h \to 0} (\underset{a}{\int^{q - h}} \frac{p}{x - q} d x + \underset{q + h}{\int^{b}} \frac{p}{x - q} d x) =$ $= \lim_{h \to 0} ({[p \ln ∣ x - q ∣]}_{a}^{q - h} + {[p \ln ∣ x - q ∣]}_{q + h}^{b}) =$ $= \lim_{h \to 0} (p \ln ∣ h ∣ - p \ln ∣ a - q ∣ + p \ln ∣ b - q ∣ - p \ln ∣ h ∣) =$ $= \lim_{h \to 0} (p \ln ∣ b - q ∣ - p \ln ∣ a - q ∣) =$ $= p \ln ∣ b - q ∣ - p \ln ∣ a - q ∣= \underset{a}{\int^{b}} \frac{p}{x - q} d x$ $q . e . d .$
	Commented by ajfour last updated on 14/May/18

	$n o t a s s u r e, y o u k n o w b e t t e r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum