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Question Number 118419 by bramlexs22 last updated on 17/Oct/20
∫ ((2sin 2x)/(4cos x+sin 2x)) dx
$$\:\:\:\int\:\frac{\mathrm{2sin}\:\mathrm{2}{x}}{\mathrm{4cos}\:{x}+\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\: \\ $$
Commented by bramlexs22 last updated on 17/Oct/20
yes...thank you all master
$${yes}…{thank}\:{you}\:{all}\:{master}\: \\ $$
Answered by peter frank last updated on 17/Oct/20
∫ ((4sin xcos x)/(4cos x+2sinx cos x)) dx ∫((2sin x)/(2+sin x)) .....
$$\:\int\:\frac{\mathrm{4sin}\:{x}\mathrm{cos}\:{x}}{\mathrm{4cos}\:{x}+\mathrm{2sin}{x}\:\mathrm{cos}\:{x}}\:{dx} \\ $$$$\int\frac{\mathrm{2sin}\:{x}}{\mathrm{2}+\mathrm{sin}\:{x}} \\ $$$$….. \\ $$
Commented by TANMAY PANACEA last updated on 17/Oct/20
sinx=((2t)/(1+t^2 )) t=tan(x/2)→2dt=sec^2 (x/2)dt ∫((2×((2t)/(1+t^2 ))×((2dt)/(1+t^2 )))/(2+((2t)/(1+t^2 )))) ∫((8tdt)/((1+t^2 )^2 ))×((1+t^2 )/(2+2t^2 +2t)) ∫((4t(1+t^2 ))/((1+t^2 )^2 (1+t+t^2 )))dt 4∫((tdt)/((1+t^2 )(1+t+t^2 ))) 4∫(((1+t+t^2 )−(1+t^2 ))/((1+t^2 )(1+t+t^2 )))dt 4∫(dt/(1+t^2 ))−4∫(dt/(t^2 +2×t×(1/2)+(1/4)+(3/4))) 4∫(dt/(1+t^2 ))−4∫(dt/((t+(1/2))^2 +(((√3)/2))^2 )) 4tan^(−1) (t)−4×(1/((((√3)/2))))tan^(−1) (((t+(1/2))/((√3)/2)))+c put t=tan(x/2)
$${sinx}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:\:\:{t}={tan}\frac{{x}}{\mathrm{2}}\rightarrow\mathrm{2}{dt}={sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dt} \\ $$$$\int\frac{\mathrm{2}×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }} \\ $$$$\int\frac{\mathrm{8}{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }×\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}{t}} \\ $$$$\int\frac{\mathrm{4}{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\mathrm{4}\int\frac{{tdt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{4}\int\frac{\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)−\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\mathrm{4}\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{4}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}×{t}×\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\mathrm{4}\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{4}\int\frac{{dt}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{4}{tan}^{−\mathrm{1}} \left({t}\right)−\mathrm{4}×\frac{\mathrm{1}}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{tan}^{−\mathrm{1}} \left(\frac{{t}+\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)+{c} \\ $$$${put}\:{t}={tan}\frac{{x}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by peter frank last updated on 18/Oct/20
thank you all
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{all} \\ $$
Answered by Dwaipayan Shikari last updated on 17/Oct/20
2∫((sinx)/(2+sinx))dx 2∫1−(2/(2+sinx))=2x−8∫(1/(2+((2t)/(1+t^2 )))).(1/(1+t^2 ))dt ( t=tan(x/2)) =2x−4∫(1/(1+t^2 +t))dt =2x−4∫(1/((t+(1/2))^2 +(((√3)/2))^2 ))dt =2x−(8/( (√3)))tan^(−1) ((2t+1)/( (√3))) +C =2x−(8/( (√3)))tan^(−1) ((2tan(x/2)+1)/( (√3)))+C
$$\mathrm{2}\int\frac{{sinx}}{\mathrm{2}+{sinx}}{dx} \\ $$$$\mathrm{2}\int\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}+{sinx}}=\mathrm{2}{x}−\mathrm{8}\int\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\:\left(\:{t}={tan}\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{x}−\mathrm{4}\int\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} +{t}}{dt} \\ $$$$=\mathrm{2}{x}−\mathrm{4}\int\frac{\mathrm{1}}{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{dt} \\ $$$$=\mathrm{2}{x}−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\:\:+{C} \\ $$$$=\mathrm{2}{x}−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}+{C} \\ $$
Answered by benjo_mathlover last updated on 17/Oct/20
solve ∫ ((2sin 2x)/(4cos x+sin 2x)) dx. Solution: I=∫ ((4sin xcos x)/(4cos x+2sin xcos x)) dx = ∫ ((2sin xcos x)/(2cos x+sin xcos x)) dx I =∫ ((2sin x)/(2+sin x)) dx ; let tan ((x/2)) = z ⇒(1/2)sec^2 ((x/2)) dx = dz or dx = 2cos^2 ((x/2)) dz I=2∫ [ (((2z)/(1+z^2 ))/((2z^2 +2z+2)/(1+z^2 )))] ((2/(1+z^2 )))dz = 2∫ [ ((2z)/((z^2 +z+1)(z^2 +1))) ] dz ((2z)/((z^2 +z+1)(z^2 +1))) = ((az+b)/(z^2 +z+1)) + ((cz+d)/(z^2 +1)) ⇒2z = (z^2 +1)(az+b)+(cz+d)(z^2 +z+1) z=0⇒0 = b +d ; d =−b z=1⇒2=2a+2b+3c+3d ; 2=2a−b+3c z=−1⇒−2=−2a+2b−c+d ; −2=−2a+b−c z=−2⇒−4=−10a+5b−6c+3d ; −2=−5a+b−3c a=0; b=−2 ; c=0 ; d=2 I= 2 [−2∫ (dz/(z^2 +z+1)) + 2∫ (dz/(z^2 +1)) ] I= 4 tan^(−1) (z)−4∫ (dz/((z+(1/2))^2 +(((√3)/2))^2 )) second integral let z+(1/2) = ((√3)/2) tan r ⇒r = tan^(−1) (((2z+1)/( (√3)))) I_2 =−4.((√3)/2)∫ ((sec^2 r dr)/((3/4)sec^2 r)) = −(8/( (√3))) r +c I_2 =−(8/( (√3))) tan^(−1) (((2z+1)/3))+c Thus I= 4tan^(−1) (z)−(8/( (√3))) tan^(−1) (((2z+1)/3))+c I=4tan^(−1) (tan ((x/2)))−(8/( (√3))) tan^(−1) (((2tan ((x/2))+1)/3)) + c or I = 2x −(8/( (√3))) tan^(−1) (((2tan ((x/2))+1)/( (√3)))) + c
$$\:{solve}\:\int\:\frac{\mathrm{2sin}\:\mathrm{2}{x}}{\mathrm{4cos}\:{x}+\mathrm{sin}\:\mathrm{2}{x}}\:{dx}. \\ $$$${Solution}: \\ $$$$\:{I}=\int\:\frac{\mathrm{4sin}\:{x}\mathrm{cos}\:{x}}{\mathrm{4cos}\:{x}+\mathrm{2sin}\:{x}\mathrm{cos}\:{x}}\:{dx}\:=\:\int\:\frac{\mathrm{2sin}\:{x}\mathrm{cos}\:{x}}{\mathrm{2cos}\:{x}+\mathrm{sin}\:{x}\mathrm{cos}\:{x}}\:{dx} \\ $$$$\:{I}\:=\int\:\frac{\mathrm{2sin}\:{x}}{\mathrm{2}+\mathrm{sin}\:{x}}\:{dx}\:;\:{let}\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)\:=\:{z}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\:{dx}\:=\:{dz} \\ $$$${or}\:{dx}\:=\:\mathrm{2cos}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\:{dz}\: \\ $$$${I}=\mathrm{2}\int\:\left[\:\frac{\frac{\mathrm{2}{z}}{\mathrm{1}+{z}^{\mathrm{2}} }}{\frac{\mathrm{2}{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{2}}{\mathrm{1}+{z}^{\mathrm{2}} }}\right]\:\left(\frac{\mathrm{2}}{\mathrm{1}+{z}^{\mathrm{2}} }\right){dz}\:=\:\mathrm{2}\int\:\left[\:\frac{\mathrm{2}{z}}{\left({z}^{\mathrm{2}} +{z}+\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)}\:\right]\:{dz} \\ $$$$\frac{\mathrm{2}{z}}{\left({z}^{\mathrm{2}} +{z}+\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)}\:=\:\frac{{az}+{b}}{{z}^{\mathrm{2}} +{z}+\mathrm{1}}\:+\:\frac{{cz}+{d}}{{z}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}{z}\:=\:\left({z}^{\mathrm{2}} +\mathrm{1}\right)\left({az}+{b}\right)+\left({cz}+{d}\right)\left({z}^{\mathrm{2}} +{z}+\mathrm{1}\right) \\ $$$${z}=\mathrm{0}\Rightarrow\mathrm{0}\:=\:{b}\:+{d}\:;\:{d}\:=−{b} \\ $$$${z}=\mathrm{1}\Rightarrow\mathrm{2}=\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{3}{c}+\mathrm{3}{d}\:;\:\mathrm{2}=\mathrm{2}{a}−{b}+\mathrm{3}{c} \\ $$$${z}=−\mathrm{1}\Rightarrow−\mathrm{2}=−\mathrm{2}{a}+\mathrm{2}{b}−{c}+{d}\:;\:−\mathrm{2}=−\mathrm{2}{a}+{b}−{c} \\ $$$${z}=−\mathrm{2}\Rightarrow−\mathrm{4}=−\mathrm{10}{a}+\mathrm{5}{b}−\mathrm{6}{c}+\mathrm{3}{d}\:;\:−\mathrm{2}=−\mathrm{5}{a}+{b}−\mathrm{3}{c} \\ $$$${a}=\mathrm{0};\:{b}=−\mathrm{2}\:;\:{c}=\mathrm{0}\:;\:{d}=\mathrm{2} \\ $$$${I}=\:\mathrm{2}\:\left[−\mathrm{2}\int\:\frac{{dz}}{{z}^{\mathrm{2}} +{z}+\mathrm{1}}\:+\:\mathrm{2}\int\:\frac{{dz}}{{z}^{\mathrm{2}} +\mathrm{1}}\:\right] \\ $$$${I}=\:\mathrm{4}\:\mathrm{tan}^{−\mathrm{1}} \left({z}\right)−\mathrm{4}\int\:\frac{{dz}}{\left({z}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${second}\:{integral}\:{let}\:{z}+\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{tan}\:{r}\:\Rightarrow{r}\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{z}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${I}_{\mathrm{2}} =−\mathrm{4}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} {r}\:{dr}}{\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sec}\:^{\mathrm{2}} {r}}\:=\:−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\:{r}\:+{c} \\ $$$${I}_{\mathrm{2}} =−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{z}+\mathrm{1}}{\mathrm{3}}\right)+{c} \\ $$$${Thus}\:{I}=\:\mathrm{4tan}^{−\mathrm{1}} \left({z}\right)−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{z}+\mathrm{1}}{\mathrm{3}}\right)+{c} \\ $$$${I}=\mathrm{4tan}^{−\mathrm{1}} \left(\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)\right)−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2tan}\:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}}{\mathrm{3}}\right)\:+\:{c}\: \\ $$$${or}\:{I}\:=\:\mathrm{2}{x}\:−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2tan}\:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+\:{c} \\ $$$$ \\ $$
Answered by 1549442205PVT last updated on 17/Oct/20
∫ ((2sin 2x)/(4cos x+sin 2x)) dx =∫((4sinxcosxdx)/(2cosx(2+sinx))) =∫((2sinxdx)/(2+sinx))=∫(2−(4/(2+sinx)))dx =2x−4∫(dx/(2+sinx)).(1)Put tan(x/2)=t ⇒dt=(1/2)(1+t^2 )dx ∫(dx/(2+sinx))=∫((2dt)/((1+t^2 )(2+((2t)/(1+t^2 ))))) =∫((2dt)/(2t^2 +2t+2))=∫(dt/(t^2 +t+1))=∫(dt/((t+(1/2))^2 +(((√3)/2))^2 )) =(2/( (√3)))tan^− (((2t+1)/( (√3))))(2).Since ∫(dx/(x^2 +a^2 ))=(1/a)tan^(−1) ((x/a)) From(1)(2)we get: ∫ ((2sin 2x)/(4cos x+sin 2x)) dx =2x−(8/( (√3)))tan^(−1) (((2tan(x/2)+1)/( (√3))))+C
$$\:\:\:\int\:\frac{\mathrm{2sin}\:\mathrm{2}{x}}{\mathrm{4cos}\:{x}+\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\:=\int\frac{\mathrm{4sinxcosxdx}}{\mathrm{2cosx}\left(\mathrm{2}+\mathrm{sinx}\right)} \\ $$$$=\int\frac{\mathrm{2sinxdx}}{\mathrm{2}+\mathrm{sinx}}=\int\left(\mathrm{2}−\frac{\mathrm{4}}{\mathrm{2}+\mathrm{sinx}}\right)\mathrm{dx} \\ $$$$=\mathrm{2x}−\mathrm{4}\int\frac{\mathrm{dx}}{\mathrm{2}+\mathrm{sinx}}.\left(\mathrm{1}\right)\mathrm{Put}\:\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}=\mathrm{t} \\ $$$$\Rightarrow\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\int\frac{\mathrm{dx}}{\mathrm{2}+\mathrm{sinx}}=\int\frac{\mathrm{2dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{2}+\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)} \\ $$$$=\int\frac{\mathrm{2dt}}{\mathrm{2t}^{\mathrm{2}} +\mathrm{2t}+\mathrm{2}}=\int\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{1}}=\int\frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−} \left(\frac{\mathrm{2t}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\left(\mathrm{2}\right).\mathrm{Since}\:\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{a}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{a}}\right) \\ $$$$\mathrm{From}\left(\mathrm{1}\right)\left(\mathrm{2}\right)\mathrm{we}\:\mathrm{get}: \\ $$$$\:\:\:\int\:\frac{\mathrm{2sin}\:\mathrm{2}{x}}{\mathrm{4cos}\:{x}+\mathrm{sin}\:\mathrm{2}{x}}\:{dx}\: \\ $$$$=\mathrm{2x}−\frac{\mathrm{8}}{\:\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2tan}\frac{\mathrm{x}}{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+\mathrm{C} \\ $$

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