calculate ∫_0 ^π ((cos^2 x)/(2+3sin(2x)))dx


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Question Number 53080 by Abdo msup. last updated on 17/Jan/19

$c a l c u l a t e \int_{0}^{π} \frac{{c o s}^{2} x}{2 + 3 s i n (2 x)} d x$
Commented by maxmathsup by imad last updated on 17/Jan/19
$let A =∫_0 ^π ((cos^2 x)/(2+3sin(2x)))dx ⇒A=∫_0 ^π ((1+cos(2x))/(2(2+3sin(2x))))dx =_(2x=t) ∫_0 ^(2π) ((1+cos(t))/(2(2+3sin(t)))) (dt/2) =(1/4) ∫_0 ^(2π) ((1+cos(t))/(2+3sin(t)))dt ⇒ 4A =∫_0 ^(2π) ((1+cost)/(2+3sint)) dt = Re( ∫_0 ^(2π) ((1+e^(it) )/(2+3sint))dt) let changement e^(it) =z give ∫_0 ^(2π) ((1+e^(it) )/(2+3sint))dt =∫_(∣z∣=1) ((1+z)/(2+3((z−z^(−1) )/(2i))))(dz/(iz)) = ∫_(∣z∣=1) ((2i(1+z))/(iz(4i +3z−3z^(−1) )))dz =∫_(∣z∣=1) ((2(1+z))/(4iz +3z^2 −3))dz let consider the complex function ϕ(z)=((2(1+z))/(3z^2 +4iz −3)) poles of ϕ? let find roots of 3z^2 +4iz −3 Δ^′ =(2i)^2 −3(−3) =−4+9 =5 ⇒z_1 =((−2i+(√5))/3) and z_2 =((−2i−(√5))/3)(are poles of ϕ) ∣z_1 ∣−1 =(1/3)(√(4+5))=3 >0 ⇒z_1 is out of circle ⇒Res(ϕ,z_1 )=0 ∣z_2 ∣−1 =(1/3)(√(4+5))=3>0 ⇒z_2 is out of circle ⇒Res(ϕ,z_2 )=0 ∫_(∣z∣=1) ϕ(z)dz =2iπ{ Res(ϕ,z_1 ) +Res(ϕ,z_2 )}=0 ⇒Re(∫ ϕ(zdz)=0 ⇒ A =0 .$
$l e t A = \int_{0}^{π} \frac{{c o s}^{2} x}{2 + 3 s i n (2 x)} d x \Rightarrow A = \int_{0}^{π} \frac{1 + c o s (2 x)}{2 (2 + 3 s i n (2 x))} d x$ $=_{2 x = t} \int_{0}^{2 π} \frac{1 + c o s (t)}{2 (2 + 3 s i n (t))} \frac{d t}{2} = \frac{1}{4} \int_{0}^{2 π} \frac{1 + c o s (t)}{2 + 3 s i n (t)} d t \Rightarrow$ $4 A = \int_{0}^{2 π} \frac{1 + c o s t}{2 + 3 s i n t} d t = R e (\int_{0}^{2 π} \frac{1 + e^{i t}}{2 + 3 s i n t} d t) l e t c h a n g e m e n t e^{i t} = z g i v e$ $\int_{0}^{2 π} \frac{1 + e^{i t}}{2 + 3 s i n t} d t = \int_{∣ z ∣= 1} \frac{1 + z}{2 + 3 \frac{z - z^{- 1}}{2 i}} \frac{d z}{i z} = \int_{∣ z ∣= 1} \frac{2 i (1 + z)}{i z (4 i + 3 z - 3 z^{- 1})} d z$ $= \int_{∣ z ∣= 1} \frac{2 (1 + z)}{4 i z + 3 z^{2} - 3} d z l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{2 (1 + z)}{3 z^{2} + 4 i z - 3} p o l e s o f φ ? l e t f i n d r o o t s o f 3 z^{2} + 4 i z - 3$ $Δ^{^{'}} = {(2 i)}^{2} - 3 (- 3) = - 4 + 9 = 5 \Rightarrow z_{1} = \frac{- 2 i + \sqrt{5}}{3} a n d z_{2} = \frac{- 2 i - \sqrt{5}}{3} (a r e p o l e s o f φ)$ $∣ z_{1} ∣ - 1 = \frac{1}{3} \sqrt{4 + 5} = 3 > 0 \Rightarrow z_{1} i s o u t o f c i r c l e \Rightarrow R e s (φ, z_{1}) = 0$ $∣ z_{2} ∣ - 1 = \frac{1}{3} \sqrt{4 + 5} = 3 > 0 \Rightarrow z_{2} i s o u t o f c i r c l e \Rightarrow R e s (φ, z_{2}) = 0$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π {R e s (φ, z_{1}) + R e s (φ, z_{2})} = 0 \Rightarrow R e (\int φ (z d z) = 0 \Rightarrow$ $A = 0 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jan/19

	$\int \frac{\frac{1 + c o s 2 x}{2}}{2 + 3 s i n 2 x} d x$ $\frac{1}{2} \int \frac{1 + c o s 2 x}{2 + 3 s i n 2 x} d x$ $\frac{1}{2} \int \frac{d x}{2 + 3 (\frac{2 t a n x}{1 + {t a n}^{2} x})} + \frac{1}{12} \int \frac{d (2 + 3 s i n 2 x)}{2 + 3 s i n 2 x}$ $\frac{1}{2} \int \frac{{s e c}^{2} x d x}{2 + 2 {t a n}^{2} x + 6 t a n x} + \frac{1}{12} l n (2 + 3 s i n 2 x) + c$ $\frac{1}{4} \int \frac{d (t a n x)}{{t a n}^{2} x + 3 t a n x + 1} + d o$ $\frac{1}{4} \int \frac{d (t a n x)}{{t a n}^{2} x + 2 t a n x \times \frac{3}{2} + \frac{9}{4} + 1 - \frac{9}{4}} + d o$ $\frac{1}{4} \int \frac{d (t a n x)}{{(t a n x + \frac{3}{2})}^{2} - {(\sqrt{\frac{5}{4}})}^{2}} + d o$ $\frac{1}{4} \int \frac{d (t a n x + \frac{3}{2})}{{(t a n x + \frac{3}{2})}^{2} - {(\sqrt{\frac{5}{4}})}^{2}} + d o$ $\frac{1}{4} \times \frac{1}{2 (\frac{\sqrt{5}}{2})} l n (\frac{t a n x + \frac{3}{2} - \frac{\sqrt{5}}{2}}{t a n x + \frac{3}{2} + \frac{\sqrt{5}}{2}}) + \frac{1}{12} l n (2 + 3 s i n 2 x) + c$ $n o w w e h a v e t o p u t l i m i t . .$ $\frac{1}{4 \sqrt{5}} ∣ l n (\frac{t a n x + \frac{3}{2} - \frac{\sqrt{5}}{2}}{t a n x + \frac{3}{2} + \frac{\sqrt{5}}{2}}) + \frac{1}{12} l n (2 + 3 s i n 2 x) ∣_{0}^{π}$ $a n s w e r i s z e r o . . .$
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