calculate ∫_0 ^π (dx/((cosx +2sinx)^2 ))


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Question Number 85801 by mathmax by abdo last updated on 24/Mar/20

$c a l c u l a t e \int_{0}^{π} \frac{d x}{{(c o s x + 2 s i n x)}^{2}}$
Commented by john santu last updated on 25/Mar/20

${(\cos x + 2 \sin x)}^{2} = 5 \cos^{2} (x - \tan^{- 1} (2))$ $\int \frac{d x}{5 \cos^{2} (x - \tan^{- 1} (2))} =$ $\frac{1}{5} \tan (x - \tan^{- 1} (2)) =$ $\frac{1}{5} (\frac{\tan x - 2}{1 + 2 \tan x}) + c$ $s o \Rightarrow \underset{0}{\int^{π}} \frac{d x}{{(\cos x + 2 \sin x)}^{2}} =$ $Missing \left or extra \right$ $\frac{1}{5} [- 2 - (- 2)] = 0$
Commented by mathmax by abdo last updated on 26/Mar/20
$et A =∫_0 ^π (dx/((cosx+2sinx)^2 )) changement tan((x/2))=t give A =∫_0 ^∞ ((2dt)/((1+t^2 ){((1−t^2 )/(1+t^2 ))+((4t)/(1+t^2 ))}^2 )) =∫_0 ^∞ ((2(1+t^2 ))/((−t^2 +4t+1)^2 ))dt =2∫_0 ^∞ ((t^2 +1)/((t^2 −4t−1)^2 ))dt let decompose F(t)=((t^2 +1)/((t^2 −4t−1)^2 )) t^2 −4t−1 =0→Δ^′ =4+1 =5 ⇒t_1 =2+(√5) and t_2 =2−(√5) F(t) =((t^2 +1)/((t−t_1 )^2 (t−t_2 )^2 )) =(a/(t−t_1 )) +(b/((t−t_1 )^2 )) +(c/(t−t_2 )) +(d/((t−t_2 )^2 )) b =((t_1 ^2 +1)/((t_1 −t_2 )^2 )) =((4+4(√5)+5+1)/((2(√5))^2 )) =((10+4(√5))/(20)) =((5+2(√5))/(10)) d =((t_2 ^2 +1)/((t_2 −t_1 )^2 )) =((4−4(√5)+5+1)/(20)) =((10−4(√5))/(20)) =((5−2(√5))/(10)) lim_(t→+∞) tF(t) =0 =a+c ⇒c=−a F(0)=(1/((t_1 t_2 )^2 )) =1 =−(a/t_1 ) +(b/t_1 ^2 ) +(a/t_2 ) +(d/t_2 ^2 ) ⇒ ((1/t_2 )−(1/t_1 ))a =1−(b/t_1 ^2 )−(d/t_2 ^2 ) ⇒(((2(√5))/(−1)))a =1−(b/(4+4(√5)+5))−(d/(4−4(√5)+5)) =1−(b/(9+4(√5)))−(d/(9−4(√5))) ⇒a =−(1/(2(√5))){1−(b/(9+4(√5)))−(d/(9−4(√5)))} ⇒ A =2a∫_0 ^∞ (dt/(t−t_1 )) +2b∫_0 ^∞ (dt/((t−t_1 )^2 )) +2c ∫_0 ^∞ (dt/(t−t_2 )) +2d∫_0 ^∞ (dt/((t−t_2 )^2 )) =[2aln∣t−t_1 ∣+2cln∣t−t_2 ∣]_0 ^(+∞) +[((−2b)/(t−t_1 ))−((2d)/(t−t_2 ))]_0 ^(+∞) rest to finish the calculus....$
$e t A = \int_{0}^{π} \frac{d x}{{(c o s x + 2 s i n x)}^{2}} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $A = \int_{0}^{\infty} \frac{2 d t}{(1 + t^{2}) {\frac{1 - t^{2}}{1 + t^{2}} + \frac{4 t}{1 + t^{2}}}^{2}} = \int_{0}^{\infty} \frac{2 (1 + t^{2})}{{(- t^{2} + 4 t + 1)}^{2}} d t$ $= 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{(t^{2} - 4 t - 1)}^{2}} d t l e t d e c o m p o s e F (t) = \frac{t^{2} + 1}{{(t^{2} - 4 t - 1)}^{2}}$ $t^{2} - 4 t - 1 = 0 \to Δ^{^{'}} = 4 + 1 = 5 \Rightarrow t_{1} = 2 + \sqrt{5} a n d t_{2} = 2 - \sqrt{5}$ $F (t) = \frac{t^{2} + 1}{{(t - t_{1})}^{2} {(t - t_{2})}^{2}} = \frac{a}{t - t_{1}} + \frac{b}{{(t - t_{1})}^{2}} + \frac{c}{t - t_{2}} + \frac{d}{{(t - t_{2})}^{2}}$ $b = \frac{t_{1}^{2} + 1}{{(t_{1} - t_{2})}^{2}} = \frac{4 + 4 \sqrt{5} + 5 + 1}{{(2 \sqrt{5})}^{2}} = \frac{10 + 4 \sqrt{5}}{20} = \frac{5 + 2 \sqrt{5}}{10}$ $d = \frac{t_{2}^{2} + 1}{{(t_{2} - t_{1})}^{2}} = \frac{4 - 4 \sqrt{5} + 5 + 1}{20} = \frac{10 - 4 \sqrt{5}}{20} = \frac{5 - 2 \sqrt{5}}{10}$ ${l i m}_{t \to + \infty} t F (t) = 0 = a + c \Rightarrow c = - a$ $F (0) = \frac{1}{{(t_{1} t_{2})}^{2}} = 1 = - \frac{a}{t_{1}} + \frac{b}{t_{1}^{2}} + \frac{a}{t_{2}} + \frac{d}{t_{2}^{2}} \Rightarrow$ $(\frac{1}{t_{2}} - \frac{1}{t_{1}}) a = 1 - \frac{b}{t_{1}^{2}} - \frac{d}{t_{2}^{2}} \Rightarrow (\frac{2 \sqrt{5}}{- 1}) a = 1 - \frac{b}{4 + 4 \sqrt{5} + 5} - \frac{d}{4 - 4 \sqrt{5} + 5}$ $= 1 - \frac{b}{9 + 4 \sqrt{5}} - \frac{d}{9 - 4 \sqrt{5}} \Rightarrow a = - \frac{1}{2 \sqrt{5}} {1 - \frac{b}{9 + 4 \sqrt{5}} - \frac{d}{9 - 4 \sqrt{5}}} \Rightarrow$ $A = 2 a \int_{0}^{\infty} \frac{d t}{t - t_{1}} + 2 b \int_{0}^{\infty} \frac{d t}{{(t - t_{1})}^{2}} + 2 c \int_{0}^{\infty} \frac{d t}{t - t_{2}} + 2 d \int_{0}^{\infty} \frac{d t}{{(t - t_{2})}^{2}}$ $= {[2 a l n ∣ t - t_{1} ∣ + 2 c l n ∣ t - t_{2} ∣]}_{0}^{+ \infty} + {[\frac{- 2 b}{t - t_{1}} - \frac{2 d}{t - t_{2}}]}_{0}^{+ \infty}$ $r e s t t o f i n i s h t h e c a l c u l u s . . . .$
Commented by mathmax by abdo last updated on 26/Mar/20
$let try another way we have A =2 ∫_0 ^∞ ((t^2 +1)/((t−t_1 )^2 (t−t_2 )^2 )) with t_1 =2+(√5) and t_2 =2−(√5) A =2∫_0 ^∞ ((t^2 +1)/((((t−t_1 )/(t−t_2 )))^2 (t−t_2 )^4 ))dt we do the changement ((t−t_1 )/(t−t_2 )) =u⇒ t−t_1 =ut−ut_2 ⇒(1−u)t =t_1 −ut_2 ⇒t =((t_1 −ut_2 )/(1−u)) ⇒ dt =((−t_2 (1−u)+(t_1 −ut_2 ))/((1−u)^2 )) =((−t_2 +t_2 u+t_1 −ut_2 )/((1−u)^2 )) =((2(√5))/((1−u)^2 )) t−t_2 =((t_1 −ut_2 )/(1−u))−t_2 =((t_1 −ut_2 −t_2 +ut_2 )/(1−u)) =((2(√5))/(1−u)) ⇒ A =2 ∫_(t_1 /t_2 ) ^1 (((((t_1 −ut_2 )/(1−u)))^2 +1)/(u^2 ×(((2(√5))^4 )/((1−u)^4 ))))×((2(√5))/((1−u)^2 ))du A =(2/((2(√5))^3 )) ∫_(t_1 /t_2 ) ^1 (((t_1 −ut_2 )^2 +(1−u)^2 )/u^2 ) du A =(1/(20(√5))) ∫_(t_1 /t_2 ) ^1 ((t_2 ^2 u^2 +2u +t_1 ^2 +u^2 −2u +1)/u^2 )du (t_1 t_2 =−1) 20(√5) A =∫_(t_1 /t_2 ) ^1 (((1+t_2 ^2 )u^2 +t_1 ^2 )/u^2 )du =(1+t_2 ^2 )(1−(t_1 /t_2 )) −t_1 ^2 [(1/u)]_(t_1 /t_2 ) ^1 =(1+t_2 ^2 )((t_2 −t_1 )/t_2 ) −t_1 ^2 (1−(t_2 /t_1 )) ⇒ A=(1/(20(√5))){(1+(2−(√5))^2 )(((−2(√5))/(2−(√5))))−(2+(√5))^2 (((2(√5))/(2+(√5))))}$
$l e t t r y a n o t h e r w a y w e h a v e$ $A = 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{(t - t_{1})}^{2} {(t - t_{2})}^{2}} w i t h t_{1} = 2 + \sqrt{5} a n d t_{2} = 2 - \sqrt{5}$ $A = 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{(\frac{t - t_{1}}{t - t_{2}})}^{2} {(t - t_{2})}^{4}} d t w e d o t h e c h a n g e m e n t \frac{t - t_{1}}{t - t_{2}} = u \Rightarrow$ $t - t_{1} = u t - {u t}_{2} \Rightarrow (1 - u) t = t_{1} - {u t}_{2} \Rightarrow t = \frac{t_{1} - {u t}_{2}}{1 - u} \Rightarrow$ $d t = \frac{- t_{2} (1 - u) + (t_{1} - {u t}_{2})}{{(1 - u)}^{2}} = \frac{- t_{2} + t_{2} u + t_{1} - {u t}_{2}}{{(1 - u)}^{2}} = \frac{2 \sqrt{5}}{{(1 - u)}^{2}}$ $t - t_{2} = \frac{t_{1} - {u t}_{2}}{1 - u} - t_{2} = \frac{t_{1} - {u t}_{2} - t_{2} + {u t}_{2}}{1 - u} = \frac{2 \sqrt{5}}{1 - u} \Rightarrow$ $A = 2 \int_{\frac{t_{1}}{t_{2}}}^{1} \frac{{(\frac{t_{1} - {u t}_{2}}{1 - u})}^{2} + 1}{u^{2} \times \frac{{(2 \sqrt{5})}^{4}}{{(1 - u)}^{4}}} \times \frac{2 \sqrt{5}}{{(1 - u)}^{2}} d u$ $A = \frac{2}{{(2 \sqrt{5})}^{3}} \int_{\frac{t_{1}}{t_{2}}}^{1} \frac{{(t_{1} - {u t}_{2})}^{2} + {(1 - u)}^{2}}{u^{2}} d u$ $A = \frac{1}{20 \sqrt{5}} \int_{\frac{t_{1}}{t_{2}}}^{1} \frac{t_{2}^{2} u^{2} + 2 u + t_{1}^{2} + u^{2} - 2 u + 1}{u^{2}} d u (t_{1} t_{2} = - 1)$ $20 \sqrt{5} A = \int_{\frac{t_{1}}{t_{2}}}^{1} \frac{(1 + t_{2}^{2}) u^{2} + t_{1}^{2}}{u^{2}} d u$ $= (1 + t_{2}^{2}) (1 - \frac{t_{1}}{t_{2}}) - t_{1}^{2} {[\frac{1}{u}]}_{\frac{t_{1}}{t_{2}}}^{1} = (1 + t_{2}^{2}) \frac{t_{2} - t_{1}}{t_{2}} - t_{1}^{2} (1 - \frac{t_{2}}{t_{1}}) \Rightarrow$ $A = \frac{1}{20 \sqrt{5}} {(1 + {(2 - \sqrt{5})}^{2}) (\frac{- 2 \sqrt{5}}{2 - \sqrt{5}}) - {(2 + \sqrt{5})}^{2} (\frac{2 \sqrt{5}}{2 + \sqrt{5}})}$
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