let a^2 >b^(2 ) +c^2 calculate ∫_0 ^(2π) (dθ/(a+bsinθ +c cosθ))


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Question Number 46850 by maxmathsup by imad last updated on 01/Nov/18

$l e t a^{2} > b^{2} + c^{2} c a l c u l a t e \int_{0}^{2 π} \frac{d θ}{a + b s i n θ + c c o s θ}$
Commented by maxmathsup by imad last updated on 01/Nov/18
$residus method let I = ∫_0 ^(2π) (dθ/(a+bsinθ +ccosθ)) changement z =e^(iθ) give I = ∫_(∣z∣=1) (1/(a +b((z−z^(−1) )/(2i))+c((z+z^(−1) )/2))) (dz/(iz)) =∫_(∣z∣=1) (dz/(iz{a+b((z−z^(−1) )/(2i))+c((z+z^(−1) )/2)})) = ∫_(∣z∣=1) ((2dz)/(z{2ia +b(z−z^(−1) )+ci(z+z^(−1) ) })) = ∫_(∣z∣=1) ((2dz)/(2iaz +bz^2 −b +ciz^2 +ci)) =∫_(∣z∣=1) ((2dz)/((b+ci)z^2 +2iaz +ci−b)) let ϕ(z) =(2/((b+ci)z^2 +2iaz +ci−b)) .poles of ϕ? Δ^′ =(ia)^2 −(b+ci)(ci−b) =−a^2 −((ci)^2 −b^2 ) =−a^2 +b^2 +c^2 =−(a^2 −(b^2 +c^2 ))<0 ⇒ z_1 =((−ia +i(√(a^2 −b^2 −c^2 )))/(b+ci)) z_2 =((−ia−i(√(a^2 −b^2 −c^2 )))/(b+ci)) ∣z_1 ∣−1 =((∣a−(√(a^2 −b^2 −c^2 ))∣)/(√(b^2 +c^2 ))) >1 ⇒z_1 is out of circle . ⇒Res(ϕ,z_1 )=0 ∣z_2 ∣−1 =((∣a+(√(a^2 −b^2 −c^2 ))∣ )/(√(b^2 +c^2 ))) >1 ⇒ z_2 is out of circle ⇒Res(ϕ,z_(2 ) )=0 ∫_0 ^(2π) ϕ(z)dz =0$
$r e s i d u s m e t h o d l e t I = \int_{0}^{2 π} \frac{d θ}{a + b s i n θ + c c o s θ} c h a n g e m e n t z = e^{i θ} g i v e$ $I = \int_{∣ z ∣= 1} \frac{1}{a + b \frac{z - z^{- 1}}{2 i} + c \frac{z + z^{- 1}}{2}} \frac{d z}{i z} = \int_{∣ z ∣= 1} \frac{d z}{i z {a + b \frac{z - z^{- 1}}{2 i} + c \frac{z + z^{- 1}}{2}}}$ $= \int_{∣ z ∣= 1} \frac{2 d z}{z {2 i a + b (z - z^{- 1}) + c i (z + z^{- 1})}}$ $= \int_{∣ z ∣= 1} \frac{2 d z}{2 i a z + {b z}^{2} - b + {c i z}^{2} + c i} = \int_{∣ z ∣= 1} \frac{2 d z}{(b + c i) z^{2} + 2 i a z + c i - b}$ $l e t φ (z) = \frac{2}{(b + c i) z^{2} + 2 i a z + c i - b} . p o l e s o f φ ?$ $Δ^{^{'}} = {(i a)}^{2} - (b + c i) (c i - b) = - a^{2} - ({(c i)}^{2} - b^{2})$ $= - a^{2} + b^{2} + c^{2} = - (a^{2} - (b^{2} + c^{2})) < 0 \Rightarrow z_{1} = \frac{- i a + i \sqrt{a^{2} - b^{2} - c^{2}}}{b + c i}$ $z_{2} = \frac{- i a - i \sqrt{a^{2} - b^{2} - c^{2}}}{b + c i}$ $∣ z_{1} ∣ - 1 = \frac{∣ a - \sqrt{a^{2} - b^{2} - c^{2}} ∣}{\sqrt{b^{2} + c^{2}}} > 1 \Rightarrow z_{1} i s o u t o f c i r c l e . \Rightarrow R e s (φ, z_{1}) = 0$ $∣ z_{2} ∣ - 1 = \frac{∣ a + \sqrt{a^{2} - b^{2} - c^{2}} ∣}{\sqrt{b^{2} + c^{2}}} > 1 \Rightarrow z_{2} i s o u t o f c i r c l e \Rightarrow R e s (φ, z_{2}) = 0$ $\int_{0}^{2 π} φ (z) d z = 0$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18

	$t = t a n \frac{θ}{2} d t = {s e c}^{2} \frac{θ}{2} \times \frac{1}{2} d θ$ $d θ = \frac{2 d t}{1 + t^{2}}$ $\int \frac{2 d t}{(1 + t^{2}) (a + b \times \frac{2 t}{1 + t^{2}} + c \times \frac{1 - t^{2}}{1 + t^{2}})}$ $\int \frac{2 d t}{(a + {a t}^{2} + 2 b t + c - {c t}^{2})}$ $2 \int \frac{d t}{t^{2} (a - c) + t (2 b) + a + c}$ $\frac{2}{a - c} \int \frac{d t}{t^{2} + 2 . t . \frac{b}{a - c} + \frac{a + c}{a - c}}$ $\frac{2}{a - c} \int \frac{d t}{{(t + \frac{b}{a - c})}^{2} + \frac{a + c}{a - c} - \frac{b^{2}}{{(a - c)}^{2}}}$ $\frac{2}{a - c} \int \frac{d t}{{(t + \frac{b}{a - c})}^{2} + \frac{a^{2} - c^{2} - b^{2}}{{(a - c)}^{2}}}$ $\frac{2}{a - c} \times \frac{1}{\sqrt{\frac{a^{2} - b^{2} - c^{2}}{{(a - c)}^{2}}}} \times {t a n}^{- 1} (\frac{t + \frac{b}{a - c}}{\sqrt{\frac{a^{2} - b^{2} - c 2}{{(a - c)}^{2}}}})$ $= \frac{2}{\sqrt{a^{2} - b^{2} - c^{2}}} {t a n}^{- 1} (\frac{t + \frac{b}{a - c}}{\sqrt{\frac{a^{2} - b^{2} - c^{2}}{{(a - c)}^{2}}}})$ $w h e n θ \to 0 t \to 0$ $θ \to 2 π t \to 0$ $t h u s v a l u e o f i n y r e g a t i o n z e r o$ $p l s c h e c k w h e r e i a m w r o n g . .$ $s i n c e a^{2} > (b^{2} + c^{2})$
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