Menu Close

3sin-x-2-2sin-x-3-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 137249 by bemath last updated on 31/Mar/21
∫ (((3sin x+2))/((2sin x+3)^2 )) dx =?
$$\int\:\frac{\left(\mathrm{3sin}\:\mathrm{x}+\mathrm{2}\right)}{\left(\mathrm{2sin}\:\mathrm{x}+\mathrm{3}\right)^{\mathrm{2}} }\:\mathrm{dx}\:=? \\ $$
Answered by bemath last updated on 31/Mar/21
let tan ((x/2))=u { ((dx=(2/(u^2 +1)) du)),((sin x=((2u)/(u^2 +1)))) :} I=∫ (((((6u+2u^2 +2)/(u^2 +1))))/((((4u+3u^2 +3)/(u^2 +1)))^2 )) ((2/(u^2 +1)))du I= ∫ ((4(u^2 +3u+1))/((3u^2 +4u+3)^2 )) du ....
$$\mathrm{let}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{u}\: \begin{cases}{\mathrm{dx}=\frac{\mathrm{2}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{du}}\\{\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{2u}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}}\end{cases} \\ $$$$\mathrm{I}=\int\:\frac{\left(\frac{\mathrm{6u}+\mathrm{2u}^{\mathrm{2}} +\mathrm{2}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\right)}{\left(\frac{\mathrm{4u}+\mathrm{3u}^{\mathrm{2}} +\mathrm{3}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} }\:\left(\frac{\mathrm{2}}{\mathrm{u}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{du} \\ $$$$\mathrm{I}=\:\int\:\frac{\mathrm{4}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{3u}+\mathrm{1}\right)}{\left(\mathrm{3u}^{\mathrm{2}} +\mathrm{4u}+\mathrm{3}\right)^{\mathrm{2}} }\:\mathrm{du} \\ $$$$…. \\ $$
Answered by Ar Brandon last updated on 31/Mar/21
I=∫((3sinx+2)/((2sinx+3)^2 ))dx=(3/2)∫((2sinx+3−(5/3))/((2sinx+3)^2 ))dx =(3/2)∫(dx/(2sinx+3))−(5/2)∫(dx/((2sinx+3)^2 )) =3∫(dt/((2((2t)/(1+t^2 ))+3)(t^2 +1)))−5∫(dt/((2((2t)/(1+t^2 ))+3)^2 (t^2 +1))) =3∫(dt/(3t^2 +4t+3))−5∫((t^2 +1)/((3t^2 +4t+3)^2 ))dt f(t)=((t^2 +1)/((3t^2 +4t+3)^2 ))=(1/3)(((3t^2 +4t+3−4t)/((3t^2 +4t+3)^2 ))) =(1/(3(3t^2 +4t+3)))−((4t)/(3(3t^2 +4t+3)^2 )) g(t)=((4t)/(3(3t^2 +4t+3)^2 ))=(1/3){(2/3)∙((6t+4)/((3t^2 +4t+3)^2 ))−(8/3)∙(1/((3t^2 +4t+3)^2 ))}
$$\mathcal{I}=\int\frac{\mathrm{3sinx}+\mathrm{2}}{\left(\mathrm{2sinx}+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dx}=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{2sinx}+\mathrm{3}−\frac{\mathrm{5}}{\mathrm{3}}}{\left(\mathrm{2sinx}+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$\:\:\:=\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{dx}}{\mathrm{2sinx}+\mathrm{3}}−\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{dx}}{\left(\mathrm{2sinx}+\mathrm{3}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:=\mathrm{3}\int\frac{\mathrm{dt}}{\left(\mathrm{2}\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\mathrm{3}\right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)}−\mathrm{5}\int\frac{\mathrm{dt}}{\left(\mathrm{2}\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\mathrm{3}\right)^{\mathrm{2}} \left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:=\mathrm{3}\int\frac{\mathrm{dt}}{\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}}−\mathrm{5}\int\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}\right)^{\mathrm{2}} }\mathrm{dt} \\ $$$$\mathrm{f}\left(\mathrm{t}\right)=\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}−\mathrm{4t}}{\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}\right)}−\frac{\mathrm{4t}}{\mathrm{3}\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{4t}}{\mathrm{3}\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{3}}\left\{\frac{\mathrm{2}}{\mathrm{3}}\centerdot\frac{\mathrm{6t}+\mathrm{4}}{\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}\right)^{\mathrm{2}} }−\frac{\mathrm{8}}{\mathrm{3}}\centerdot\frac{\mathrm{1}}{\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{4t}+\mathrm{3}\right)^{\mathrm{2}} }\right\} \\ $$
Answered by Dwaipayan Shikari last updated on 31/Mar/21
∫((asinx+b)/((bsinx+a)^2 ))dx=τ(a,b) =(a^2 /b)∫((absin(x)+a^2 )/((absin(x)+a^2 )^2 ))+((b^2 −a^2 )/((absin(x)+a^2 )^2 ))dx =(a/b^2 )∫(1/(sin(x)+(a/b)))dx+(((b^2 −a^2 ))/b^3 )∫(1/((sin(x)+(a/b))^2 ))dx tan(x/2)=t J(a)=∫(1/(sin(x)+(a/b)))dx=2∫(1/(((2t)/(1+t^2 ))+(a/b))).(1/(1+t^2 ))dt =((2b)/a)∫(1/(t^2 +2(b/a)t+1))dt=((2b)/a)∫(1/((t+(√(b/a)))^2 +((b−a)/a)))dt =((2b)/a)(√(a/(b−a)))tan^(−1) (((√a)t+(√b))/( (√(b−a))))+C J′(a)=−(1/b)∫(1/((sinx+(a/b))^2 ))dx J′(a)=tan^(−1) (((√a)t+(√b))/( (√(b−a))))(−((b(b−a)^(−1/2) )/a^(3/2) )+(a^(−1/2) /((b−a)^(3/2) )))+((b−a)/(b−a+((√a)t+(√b))^2 ))((((√a)t+b)/( 2(b−a)^(3/2) ))+((t/( (√a)))/2)) ...τ(a,b)=(a^2 /b)J(a)−((b^2 −a^2 )/b^4 )J′(a)
$$\int\frac{{asinx}+{b}}{\left({bsinx}+{a}\right)^{\mathrm{2}} }{dx}=\tau\left({a},{b}\right) \\ $$$$=\frac{{a}^{\mathrm{2}} }{{b}}\int\frac{{absin}\left({x}\right)+{a}^{\mathrm{2}} }{\left({absin}\left({x}\right)+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }+\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\left({absin}\left({x}\right)+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{{a}}{{b}^{\mathrm{2}} }\int\frac{\mathrm{1}}{{sin}\left({x}\right)+\frac{{a}}{{b}}}{dx}+\frac{\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)}{{b}^{\mathrm{3}} }\int\frac{\mathrm{1}}{\left({sin}\left({x}\right)+\frac{{a}}{{b}}\right)^{\mathrm{2}} }{dx}\:\:\:\:\:{tan}\frac{{x}}{\mathrm{2}}={t} \\ $$$${J}\left({a}\right)=\int\frac{\mathrm{1}}{{sin}\left({x}\right)+\frac{{a}}{{b}}}{dx}=\mathrm{2}\int\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{{a}}{{b}}}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{2}{b}}{{a}}\int\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{2}\frac{{b}}{{a}}{t}+\mathrm{1}}{dt}=\frac{\mathrm{2}{b}}{{a}}\int\frac{\mathrm{1}}{\left({t}+\sqrt{\frac{{b}}{{a}}}\right)^{\mathrm{2}} +\frac{{b}−{a}}{{a}}}{dt} \\ $$$$=\frac{\mathrm{2}{b}}{{a}}\sqrt{\frac{{a}}{{b}−{a}}}{tan}^{−\mathrm{1}} \frac{\sqrt{{a}}{t}+\sqrt{{b}}}{\:\sqrt{{b}−{a}}}+{C} \\ $$$${J}'\left({a}\right)=−\frac{\mathrm{1}}{{b}}\int\frac{\mathrm{1}}{\left({sinx}+\frac{{a}}{{b}}\right)^{\mathrm{2}} }{dx} \\ $$$${J}'\left({a}\right)={tan}^{−\mathrm{1}} \frac{\sqrt{{a}}{t}+\sqrt{{b}}}{\:\sqrt{{b}−{a}}}\left(−\frac{{b}\left({b}−{a}\right)^{−\mathrm{1}/\mathrm{2}} }{{a}^{\mathrm{3}/\mathrm{2}} }+\frac{{a}^{−\mathrm{1}/\mathrm{2}} }{\left({b}−{a}\right)^{\mathrm{3}/\mathrm{2}} }\right)+\frac{{b}−{a}}{{b}−{a}+\left(\sqrt{{a}}{t}+\sqrt{{b}}\right)^{\mathrm{2}} }\left(\frac{\sqrt{{a}}{t}+{b}}{\:\mathrm{2}\left({b}−{a}\right)^{\mathrm{3}/\mathrm{2}} }+\frac{\frac{{t}}{\:\sqrt{{a}}}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$…\tau\left({a},{b}\right)=\frac{{a}^{\mathrm{2}} }{{b}}{J}\left({a}\right)−\frac{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{b}^{\mathrm{4}} }{J}'\left({a}\right) \\ $$$$ \\ $$
Answered by MJS_new last updated on 31/Mar/21
(u/v)′=((u′v−v′u)/v^2 ) ⇔ (u/v)=∫((u′)/v)−((v′u)/v^2 ) ⇔ ∫((v′u)/v^2 )=−(u/v)+∫((u′)/v) v=3+2sin x ⇒ v′=2cos x 2ucos x =2+3sin x ⇒ u=((2+3sin x)/(2cos x)) ⇒ u′=((3+2sin x)/(2cos^2 x)) now we have ∫((v′u)/v^2 )=−(u/v)+∫((u′)/v) ∫((2+3sin x)/((3+2sin x)^2 ))dx=−((2+3sin x)/(2(3+2sin x)cos x))+(1/2)∫(dx/(cos^2 x))= =−((2+3sin x)/(2(3+2sin x)cos x))+(1/2)tan x = =−((cos x)/(3+2sin x))+C
$$\left({u}/{v}\right)'=\frac{{u}'{v}−{v}'{u}}{{v}^{\mathrm{2}} }\:\Leftrightarrow\:\frac{{u}}{{v}}=\int\frac{{u}'}{{v}}−\frac{{v}'{u}}{{v}^{\mathrm{2}} }\:\Leftrightarrow\:\int\frac{{v}'{u}}{{v}^{\mathrm{2}} }=−\frac{{u}}{{v}}+\int\frac{{u}'}{{v}} \\ $$$${v}=\mathrm{3}+\mathrm{2sin}\:{x}\:\Rightarrow\:{v}'=\mathrm{2cos}\:{x} \\ $$$$\mathrm{2}{u}\mathrm{cos}\:{x}\:=\mathrm{2}+\mathrm{3sin}\:{x}\:\Rightarrow\:{u}=\frac{\mathrm{2}+\mathrm{3sin}\:{x}}{\mathrm{2cos}\:{x}}\:\Rightarrow\:{u}'=\frac{\mathrm{3}+\mathrm{2sin}\:{x}}{\mathrm{2cos}^{\mathrm{2}} \:{x}} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int\frac{{v}'{u}}{{v}^{\mathrm{2}} }=−\frac{{u}}{{v}}+\int\frac{{u}'}{{v}} \\ $$$$\int\frac{\mathrm{2}+\mathrm{3sin}\:{x}}{\left(\mathrm{3}+\mathrm{2sin}\:{x}\right)^{\mathrm{2}} }{dx}=−\frac{\mathrm{2}+\mathrm{3sin}\:{x}}{\mathrm{2}\left(\mathrm{3}+\mathrm{2sin}\:{x}\right)\mathrm{cos}\:{x}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\mathrm{cos}^{\mathrm{2}} \:{x}}= \\ $$$$=−\frac{\mathrm{2}+\mathrm{3sin}\:{x}}{\mathrm{2}\left(\mathrm{3}+\mathrm{2sin}\:{x}\right)\mathrm{cos}\:{x}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:{x}\:= \\ $$$$=−\frac{\mathrm{cos}\:{x}}{\mathrm{3}+\mathrm{2sin}\:{x}}+{C} \\ $$
Commented by Dwaipayan Shikari last updated on 31/Mar/21
So it has been proved that experience pays more :(
$${So}\:{it}\:{has}\:{been}\:{proved}\:{that}\:{experience}\:{pays}\:{more}\::\left(\right. \\ $$
Commented by MJS_new last updated on 31/Mar/21
how old are you? I′ve been solving integrals for about 35 years... this special one I tried using t=tan (x/2) first; then I found Ostrogradski′s Method leads to ((p_1 (t))/(q_1 (t)))+∫0dt so I knew it can be solved by parts using the quotient rule most people don′t know...
$$\mathrm{how}\:\mathrm{old}\:\mathrm{are}\:\mathrm{you}? \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{been}\:\mathrm{solving}\:\mathrm{integrals}\:\mathrm{for}\:\mathrm{about}\:\mathrm{35}\:\mathrm{years}… \\ $$$$\mathrm{this}\:\mathrm{special}\:\mathrm{one}\:\mathrm{I}\:\mathrm{tried}\:\mathrm{using}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{first}; \\ $$$$\mathrm{then}\:\mathrm{I}\:\mathrm{found}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{{p}_{\mathrm{1}} \left({t}\right)}{{q}_{\mathrm{1}} \left({t}\right)}+\int\mathrm{0}{dt}\:\mathrm{so}\:\mathrm{I}\:\mathrm{knew}\:\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{parts} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{quotient}\:\mathrm{rule}\:\mathrm{most}\:\mathrm{people}\:\mathrm{don}'\mathrm{t} \\ $$$$\mathrm{know}… \\ $$
Commented by Dwaipayan Shikari last updated on 31/Mar/21
e^((17)/6) (Exactly maybe) :)
$$\left.{e}^{\frac{\mathrm{17}}{\mathrm{6}}} \:\:\left({Exactly}\:{maybe}\right)\:\::\right) \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 31/Mar/21
35 years ! That′s the sum of our ages 😅
$$\mathrm{35}\:\mathrm{years}\:!\:\mathrm{That}'\mathrm{s}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{our}\:\mathrm{ages}\: \\ $$😅
Answered by EDWIN88 last updated on 31/Mar/21
E = ∫ (((2+3sin x))/((3+2sin x)^2 )) dx E = ∫ ((2+3sin x)/(2cos x)). ((2cos x)/((3+2sin x)^2 )) dx E = (1/2)∫ ((2+3sin x)/(cos x))_(u) .((2cos x)/((3+2sin x)^2 )).dx_(dv) by parts ⇒du = ((3+2sin x)/(cos^2 x)) dx ; v = −(1/(3+2sin x)) 2E=−((2+3sin x)/(cos x(3+2sin x)))+∫ (dx/(cos^2 x)) 2E = −((2+3sin x)/(cos x(3+2sin x))) + tan x + c E = −((2+3sin x)/(2cos x(3+2sin x)))+ ((sin x)/(2cos x))+ C E=((−2−3sin x+sin x(3+2sin x))/(2cos x(3+2sin x))) + C E = ((2sin^2 x−2)/(2cos x(3+2sin x))) = ((−cos x)/(3+2sin x)) + C
$$\mathrm{E}\:=\:\int\:\frac{\left(\mathrm{2}+\mathrm{3sin}\:\mathrm{x}\right)}{\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)^{\mathrm{2}} }\:\mathrm{dx}\: \\ $$$$\mathrm{E}\:=\:\int\:\frac{\mathrm{2}+\mathrm{3sin}\:\mathrm{x}}{\mathrm{2cos}\:\mathrm{x}}.\:\frac{\mathrm{2cos}\:\mathrm{x}}{\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{E}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\underset{\mathrm{u}} {\underbrace{\:\frac{\mathrm{2}+\mathrm{3sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}}}\:.\underset{\mathrm{dv}} {\underbrace{\frac{\mathrm{2cos}\:\mathrm{x}}{\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)^{\mathrm{2}} }.\mathrm{dx}}} \\ $$$$\mathrm{by}\:\mathrm{parts}\: \\ $$$$\Rightarrow\mathrm{du}\:=\:\frac{\mathrm{3}+\mathrm{2sin}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:;\:\mathrm{v}\:=\:−\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2sin}\:\mathrm{x}} \\ $$$$\mathrm{2E}=−\frac{\mathrm{2}+\mathrm{3sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)}+\int\:\frac{\mathrm{dx}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\: \\ $$$$\mathrm{2E}\:=\:−\frac{\mathrm{2}+\mathrm{3sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)}\:+\:\mathrm{tan}\:\mathrm{x}\:+\:\mathrm{c}\: \\ $$$$\mathrm{E}\:=\:−\frac{\mathrm{2}+\mathrm{3sin}\:\mathrm{x}}{\mathrm{2cos}\:\mathrm{x}\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)}+\:\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{2cos}\:\mathrm{x}}+\:\mathrm{C} \\ $$$$\mathrm{E}=\frac{−\mathrm{2}−\mathrm{3sin}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)}{\mathrm{2cos}\:\mathrm{x}\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)}\:+\:\mathrm{C} \\ $$$$\mathrm{E}\:=\:\frac{\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2}}{\mathrm{2cos}\:\mathrm{x}\left(\mathrm{3}+\mathrm{2sin}\:\mathrm{x}\right)}\:=\:\frac{−\mathrm{cos}\:\mathrm{x}}{\mathrm{3}+\mathrm{2sin}\:\mathrm{x}}\:+\:\mathrm{C} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *