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Question Number 128956 by mathmax by abdo last updated on 11/Jan/21
calculate ∫_0 ^π ((sin(2x))/(3+cos(4x)))dx
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{3}+\mathrm{cos}\left(\mathrm{4x}\right)}\mathrm{dx} \\ $$
Answered by chengulapetrom last updated on 11/Jan/21
cos4x=2cos^2 2x−1 ∫_0 ^π ((sin(2x))/(3+cos(4x)))dx=∫_0 ^π ((sin(2x))/(2+2cos^2 2x))dx =−(1/4)∫_0 ^π ((−2sin(2x))/(1+cos^2 2x))dx let t=cos2x⇒dt=−2sin2xdx x=0⇒t=1 and x=π⇒t=1 ∫_1 ^1 (dt/(1+t^2 ))=0
$${cos}\mathrm{4}{x}=\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{x}−\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{3}+{cos}\left(\mathrm{4}{x}\right)}{dx}=\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}+\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{x}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi} \frac{−\mathrm{2}{sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{cos}^{\mathrm{2}} \mathrm{2}{x}}{dx} \\ $$$${let}\:{t}={cos}\mathrm{2}{x}\Rightarrow{dt}=−\mathrm{2}{sin}\mathrm{2}{xdx} \\ $$$${x}=\mathrm{0}\Rightarrow{t}=\mathrm{1}\:{and}\:{x}=\pi\Rightarrow{t}=\mathrm{1} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{0} \\ $$
Commented by chengulapetrom last updated on 11/Jan/21
am I correct?
$${am}\:{I}\:{correct}? \\ $$
Answered by Olaf last updated on 11/Jan/21
Ω = ∫_0 ^π ((sin(2x))/(3+cos(4x)))dx Ω = ∫_0 ^π ((sin(2x))/(3+2cos^2 (2x)−1))dx Ω = −(1/4)∫_0 ^π ((−2sin(2x))/(1+cos^2 (2x)))dx Ω = −(1/4)[arctan(cos(2x))]_0 ^π = 0
$$\Omega\:=\:\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{2}{x}\right)}{\mathrm{3}+\mathrm{cos}\left(\mathrm{4}{x}\right)}{dx} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sin}\left(\mathrm{2}{x}\right)}{\mathrm{3}+\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{1}}{dx} \\ $$$$\Omega\:=\:−\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\pi} \frac{−\mathrm{2sin}\left(\mathrm{2}{x}\right)}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)}{dx} \\ $$$$\Omega\:=\:−\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{arctan}\left(\mathrm{cos}\left(\mathrm{2}{x}\right)\right)\right]_{\mathrm{0}} ^{\pi} \:=\:\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 11/Jan/21
let I=∫_0 ^π ((sin(2x))/(3+cos(4x)))dx ⇒I=_(2x=t) ∫_0 ^(2π) ((sint)/(3+cos(2t)))(dt/2) =(1/2)∫_0 ^(2π) ((sint)/(3+1−2sin^2 t))dt =(1/2)∫_0 ^(2π) ((sint)/(4−2sin^2 t))dt =(1/4)∫_0 ^(2π) ((sint)/(2−sin^2 t))dt ⇒4I =∫_0 ^π ((sint)/(2−sin^2 t))dt +∫_π ^(2π) ((sint)/(2−sin^2 t))dt(→t=x+π =∫_0 ^π ((sint)/(2−sin^2 t)) −∫_0 ^π ((sinx)/(2−sin^2 x))dx =0 ⇒4I=0 ⇒I=0
$$\mathrm{let}\:\mathrm{I}=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\mathrm{3}+\mathrm{cos}\left(\mathrm{4x}\right)}\mathrm{dx}\:\Rightarrow\mathrm{I}=_{\mathrm{2x}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{sint}}{\mathrm{3}+\mathrm{cos}\left(\mathrm{2t}\right)}\frac{\mathrm{dt}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{sint}}{\mathrm{3}+\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{t}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{sint}}{\mathrm{4}−\mathrm{2sin}^{\mathrm{2}} \mathrm{t}}\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{sint}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\mathrm{dt} \\ $$$$\Rightarrow\mathrm{4I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{sint}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\mathrm{dt}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\frac{\mathrm{sint}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\mathrm{dt}\left(\rightarrow\mathrm{t}=\mathrm{x}+\pi\right. \\ $$$$=\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{sint}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\:−\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{sinx}}{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}\:=\mathrm{0}\:\Rightarrow\mathrm{4I}=\mathrm{0}\:\Rightarrow\mathrm{I}=\mathrm{0} \\ $$

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