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


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Question Number 93265 by abdomathmax last updated on 12/May/20

$c a l c u l a t e \int_{0}^{\frac{2 π}{3}} \frac{d x}{3 + s i n (3 x)}$
Commented by mathmax by abdo last updated on 12/May/20

$A = \int_{0}^{\frac{2 π}{3}} \frac{d x}{3 + s i n (3 x)} \Rightarrow A =_{3 x = t} \frac{1}{3} \int_{0}^{2 π} \frac{d t}{3 + s i n t}$ $=_{e^{i t} = z} \frac{1}{3} \int_{∣ z ∣= 1} \frac{1}{3 + \frac{z - z^{- 1}}{2 i}} \times \frac{d z}{i z}$ $3 A = \int_{∣ z ∣= 1} \frac{2 i d z}{i z (6 i + z - z^{- 1})} = \int_{∣ z ∣= 1} \frac{2 d z}{6 i z + z^{2} - 1}$ $l e t φ (z) = \frac{2}{z^{2} + 6 i z - 1} p o l e s o f φ ?$ $z^{2} + 6 i z - 1 = 0 \to Δ^{^{'}} = {(3 i)}^{2} + 1 = - 8 \Rightarrow z_{1} = - 3 i + 2 i \sqrt{2}$ $z_{2} = - 3 i - 2 i \sqrt{2} w e h a v e ∣ z_{1} ∣ - 1 =∣ 3 - \sqrt{2} ∣ - 1 = 3 - \sqrt{2} - 1 = 2 - \sqrt{2} > 0$ $∣ z_{2} ∣ - 1 = 3 + \sqrt{2} - 1 = 2 + \sqrt{2} > 0 \Rightarrow R e s (φ, z_{1}) = R e s (φ, z_{2}) = 0 \Rightarrow$ $Σ R e s = 0 \Rightarrow A = 0$
	Answered by niroj last updated on 12/May/20
	$∫_0 ^( ((2π)/3)) (( dx)/(3 + sin (3x))) Put , 3x= t 3dx=dt dx=(1/3)dt IF x=((2π)/3) ⇒ t= 2π IF x=0 ⇒ t=0 (1/3)∫_0 ^( 2π) (( 1)/(3+sin t))dt =[ (1/3)∫ (( 1)/(3+ ((2tan(t/2))/(1+tan^2 (t/2)))))dt]_0 2π = [ (1/3)∫(( sec^2 (t/2))/(3+3tan^2 (t/2)+2tan(t/2)))dt]_0 ^(2π) Put, tan (t/2)= m sec^2 (t/2)dt=2dm [ (1/3)∫ ((2dm)/(3+3m^2 +2m))]_0 ^(2π) = [(2/3)∫ (1/(3(m^2 +(2/3)m+1)))dm]_0 ^(2π) = [ (2/9)∫ (( 1)/((m)^2 +2.(1/3).m+(1/9)−(1/9)+1))dm]_0 ^(2π) = [ (2/9)∫ (( 1)/((m+(1/3))^2 −((8/9))))dm]_0 ^(2π) = [ (2/9)∫ (1/((m+(1/3))^2 −(((√8)/3))^2 ))dm]_0 ^(2π) = [(2/9).(1/(2.((√8)/3)))log (( m+(1/3) +((√8)/3))/(m+(1/3) −((√8)/3)))]_0 ^(2π) =[ (1/(3(√8))) log ((3m+1+(√8))/(3m+1−(√8)))]_0 ^(2π) = [(1/(3(√8)))log ((3 tan(t/2) +1+(√8))/(3tan(t/2) +1−(√8)))]_0 ^(2π) = {(1/(3(√8)))log ((3tan ((2π)/2)+1+(√8))/(3tan((2π)/2)+1−(√8)))}−{(1/(3(√8)))log ((3.0+1+(√8))/(3.0+1−(√8)))} = (1/(3(√8))) log ((1+(√8))/(1−(√8))) −(1/(3(√8)))log ((1+(√8))/(1−(√8))) = 0 //.$
	$\int_{0}^{\frac{2 π}{3}} \frac{dx}{3 + \sin (3 x)}$ $Put, 3 x = t$ $3 dx = dt$ $dx = \frac{1}{3} dt$ $IF x = \frac{2 π}{3} \Rightarrow t = 2 π$ $IF x = 0 \Rightarrow t = 0$ $\frac{1}{3} \int_{0}^{2 π} \frac{1}{3 + \sin t} dt$ $= {[\frac{1}{3} \int \frac{1}{3 + \frac{2 \tan \frac{t}{2}}{1 + \tan^{2} \frac{t}{2}}} dt]}_{0} 2 π$ $= {[\frac{1}{3} \int \frac{\sec^{2} \frac{t}{2}}{3 + {3 \tan}^{2} \frac{t}{2} + 2 \tan \frac{t}{2}} dt]}_{0}^{2 π}$ $Put, \tan \frac{t}{2} = m$ $\sec^{2} \frac{t}{2} dt = 2 dm$ ${[\frac{1}{3} \int \frac{2 dm}{3 + {3 m}^{2} + 2 m}]}_{0}^{2 π}$ $= {[\frac{2}{3} \int \frac{1}{3 (m^{2} + \frac{2}{3} m + 1)} dm]}_{0}^{2 π}$ $= {[\frac{2}{9} \int \frac{1}{{(m)}^{2} + 2 . \frac{1}{3} . m + \frac{1}{9} - \frac{1}{9} + 1} dm]}_{0}^{2 π}$ $= {[\frac{2}{9} \int \frac{1}{{(m + \frac{1}{3})}^{2} - (\frac{8}{9})} dm]}_{0}^{2 π}$ $= {[\frac{2}{9} \int \frac{1}{{(m + \frac{1}{3})}^{2} - {(\frac{\sqrt{8}}{3})}^{2}} dm]}_{0}^{2 π}$ $= {[\frac{2}{9} . \frac{1}{2 . \frac{\sqrt{8}}{3}} \log \frac{m + \frac{1}{3} + \frac{\sqrt{8}}{3}}{m + \frac{1}{3} - \frac{\sqrt{8}}{3}}]}_{0}^{2 π}$ $= {[\frac{1}{3 \sqrt{8}} \log \frac{3 m + 1 + \sqrt{8}}{3 m + 1 - \sqrt{8}}]}_{0}^{2 π}$ $= {[\frac{1}{3 \sqrt{8}} \log \frac{3 \tan \frac{t}{2} + 1 + \sqrt{8}}{3 \tan \frac{t}{2} + 1 - \sqrt{8}}]}_{0}^{2 π}$ $= {\frac{1}{3 \sqrt{8}} \log \frac{3 \tan \frac{2 π}{2} + 1 + \sqrt{8}}{3 \tan \frac{2 π}{2} + 1 - \sqrt{8}}} - {\frac{1}{3 \sqrt{8}} \log \frac{3.0 + 1 + \sqrt{8}}{3.0 + 1 - \sqrt{8}}}$ $= \frac{1}{3 \sqrt{8}} \log \frac{1 + \sqrt{8}}{1 - \sqrt{8}} - \frac{1}{3 \sqrt{8}} \log \frac{1 + \sqrt{8}}{1 - \sqrt{8}}$ $= 0 / / .$
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