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Question Number 6444 by sanusihammed last updated on 27/Jun/16
I = ∫(dx/(sinx + sin2x))
$${I}\:=\:\int\frac{{dx}}{{sinx}\:+\:{sin}\mathrm{2}{x}} \\ $$
Commented by Temp last updated on 27/Jun/16
Try www.WolframAlpha.com search: “int 1/(sin(x)+sin(2x)) dx” if you want, for example, ∫_1 ^( 2) xdx: “int x dx from x=1 to 2”
$$\mathrm{Try}\:\mathrm{www}.\mathrm{WolframAlpha}.\mathrm{com} \\ $$$$\mathrm{search}: \\ $$$$“{int}\:\mathrm{1}/\left({sin}\left({x}\right)+{sin}\left(\mathrm{2}{x}\right)\right)\:{dx}'' \\ $$$$\mathrm{if}\:\mathrm{you}\:\mathrm{want},\:\mathrm{for}\:\mathrm{example},\:\int_{\mathrm{1}} ^{\:\mathrm{2}} {xdx}: \\ $$$$“{int}\:{x}\:{dx}\:{from}\:{x}=\mathrm{1}\:{to}\:\mathrm{2}'' \\ $$
Answered by Yozzii last updated on 27/Jun/16
I=∫(dx/(sinx+sin2x))=∫(dx/(sinx(1+2cosx))) Let u=tan0.5x⇒(2/(1+u^2 ))du=dx sinx=((2u)/(1+u^2 )) and cosx=((1−u^2 )/(1+u^2 )). ∴I=∫((2/(1+u^2 ))/(((2u)/(1+u^2 ))(1+((2(1−u^2 ))/(1+u^2 )))))du I=∫(((1+u^2 ))/(u(1+u^2 +2−2u^2 )))du I=∫((1+u^2 )/(u(3−u^2 )))du I=−∫((1+u^2 )/(u(u−(√3))(u+(√3))))du Let ((1+u^2 )/(u(u−(√3))(u+(√3))))≡(a/u)+(b/(u−(√3)))+(c/(u+(√3))) ≡((a(u^2 −3)+bu(u+(√3))+cu(u−(√3)))/(u(u+(√3))(u−(√3)))) ∴1+u^2 =a(u^2 −3)+bu(u+(√3))+cu(u−(√3)) Let u=0∴ 1=a(−3)⇒a=((−1)/3). Let u=(√3)⇒4=b(√3)(2(√3)) 4=6b⇒b=(2/3) Let u=−(√3)⇒4=c(−(√3))(−2(√3)) c=(2/3). ∴ I=−∫{(2/3)((1/(u+(√3)))+(1/(u−(√3))))−(1/(3u))}du I=−(1/3)∫{2((1/(u+(√3)))+(1/(u−(√3))))−(1/u)}du I=((−1)/3)[2ln∣(u+(√3))(u−(√3))∣−ln∣u∣]+C I=((−1)/3)ln∣(((u^2 −3)^2 )/u)∣+C I=(1/3)ln∣(u/((u^2 −3)^2 ))∣+C I=(1/3)ln∣((tan0.5x)/((tan^2 0.5x−3)^2 ))∣+C
$${I}=\int\frac{{dx}}{{sinx}+{sin}\mathrm{2}{x}}=\int\frac{{dx}}{{sinx}\left(\mathrm{1}+\mathrm{2}{cosx}\right)} \\ $$$${Let}\:{u}={tan}\mathrm{0}.\mathrm{5}{x}\Rightarrow\frac{\mathrm{2}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}={dx} \\ $$$${sinx}=\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:{and}\:{cosx}=\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }. \\ $$$$\therefore{I}=\int\frac{\mathrm{2}/\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{2}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{\mathrm{1}+{u}^{\mathrm{2}} }\right)}{du} \\ $$$${I}=\int\frac{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{{u}\left(\mathrm{1}+{u}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}{u}^{\mathrm{2}} \right)}{du} \\ $$$${I}=\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}\left(\mathrm{3}−{u}^{\mathrm{2}} \right)}{du} \\ $$$${I}=−\int\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}\left({u}−\sqrt{\mathrm{3}}\right)\left({u}+\sqrt{\mathrm{3}}\right)}{du} \\ $$$${Let}\:\frac{\mathrm{1}+{u}^{\mathrm{2}} }{{u}\left({u}−\sqrt{\mathrm{3}}\right)\left({u}+\sqrt{\mathrm{3}}\right)}\equiv\frac{{a}}{{u}}+\frac{{b}}{{u}−\sqrt{\mathrm{3}}}+\frac{{c}}{{u}+\sqrt{\mathrm{3}}} \\ $$$$\equiv\frac{{a}\left({u}^{\mathrm{2}} −\mathrm{3}\right)+{bu}\left({u}+\sqrt{\mathrm{3}}\right)+{cu}\left({u}−\sqrt{\mathrm{3}}\right)}{{u}\left({u}+\sqrt{\mathrm{3}}\right)\left({u}−\sqrt{\mathrm{3}}\right)} \\ $$$$\therefore\mathrm{1}+{u}^{\mathrm{2}} ={a}\left({u}^{\mathrm{2}} −\mathrm{3}\right)+{bu}\left({u}+\sqrt{\mathrm{3}}\right)+{cu}\left({u}−\sqrt{\mathrm{3}}\right) \\ $$$${Let}\:{u}=\mathrm{0}\therefore\:\mathrm{1}={a}\left(−\mathrm{3}\right)\Rightarrow{a}=\frac{−\mathrm{1}}{\mathrm{3}}. \\ $$$${Let}\:{u}=\sqrt{\mathrm{3}}\Rightarrow\mathrm{4}={b}\sqrt{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{4}=\mathrm{6}{b}\Rightarrow{b}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${Let}\:{u}=−\sqrt{\mathrm{3}}\Rightarrow\mathrm{4}={c}\left(−\sqrt{\mathrm{3}}\right)\left(−\mathrm{2}\sqrt{\mathrm{3}}\right) \\ $$$${c}=\frac{\mathrm{2}}{\mathrm{3}}. \\ $$$$\therefore\:{I}=−\int\left\{\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{u}+\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{{u}−\sqrt{\mathrm{3}}}\right)−\frac{\mathrm{1}}{\mathrm{3}{u}}\right\}{du} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{3}}\int\left\{\mathrm{2}\left(\frac{\mathrm{1}}{{u}+\sqrt{\mathrm{3}}}+\frac{\mathrm{1}}{{u}−\sqrt{\mathrm{3}}}\right)−\frac{\mathrm{1}}{{u}}\right\}{du} \\ $$$${I}=\frac{−\mathrm{1}}{\mathrm{3}}\left[\mathrm{2}{ln}\mid\left({u}+\sqrt{\mathrm{3}}\right)\left({u}−\sqrt{\mathrm{3}}\right)\mid−{ln}\mid{u}\mid\right]+{C} \\ $$$${I}=\frac{−\mathrm{1}}{\mathrm{3}}{ln}\mid\frac{\left({u}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} }{{u}}\mid+{C} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\frac{{u}}{\left({u}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} }\mid+{C}\: \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid\frac{{tan}\mathrm{0}.\mathrm{5}{x}}{\left({tan}^{\mathrm{2}} \mathrm{0}.\mathrm{5}{x}−\mathrm{3}\right)^{\mathrm{2}} }\mid+{C} \\ $$$$ \\ $$
Commented by nburiburu last updated on 27/Jun/16
there is another way, instead universal trigonometry substitution: ∫(1/(sin x (1+2cos x))).dx.((sin x)/(sin x))= ∫((sin x)/(sin^2 x(1+2cosx))).dx= ∫((sin x)/((1−cos^2 x)(1+2cos x))).dx t=cos x dt=−sin x dx ∫((−1)/((1−t^2 )(1+2t))).dt and using rational descomposition it is solved. 1−t^2 =(1−t)(1+t) =−∫(A/(1−t))dt −∫(B/(1+t))dt −∫(C/(1+2t))dt= =A ln(1−t)−B ln(1+t)−(C/2) ln(1+2t)+c where 1=A(1+t)(1+2t)+B(1−t)(1+2t)+C(1−t^2 ) and if t=1: 1=A.2.3⇒A=(1/6) if t=−1: 1=B(−2)(−1)⇒B=(1/2) if t=−(1/2): 1=C(1−(1/4))⇒C=(4/3) so =(1/6) ln(1−cos x)−(1/2)ln(1+cos x)−(2/3)ln(1+2cos x)+c
$${there}\:{is}\:{another}\:{way},\:{instead}\:{universal}\:{trigonometry}\:{substitution}: \\ $$$$\int\frac{\mathrm{1}}{{sin}\:{x}\:\left(\mathrm{1}+\mathrm{2}{cos}\:{x}\right)}.{dx}.\frac{{sin}\:{x}}{{sin}\:{x}}= \\ $$$$\int\frac{{sin}\:{x}}{{sin}^{\mathrm{2}} {x}\left(\mathrm{1}+\mathrm{2}{cosx}\right)}.{dx}= \\ $$$$\int\frac{{sin}\:{x}}{\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)\left(\mathrm{1}+\mathrm{2}{cos}\:{x}\right)}.{dx} \\ $$$${t}={cos}\:{x} \\ $$$${dt}=−{sin}\:{x}\:{dx} \\ $$$$\int\frac{−\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}{t}\right)}.{dt} \\ $$$${and}\:{using}\:{rational}\:{descomposition}\:{it}\:{is}\:{solved}. \\ $$$$\mathrm{1}−{t}^{\mathrm{2}} =\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+{t}\right) \\ $$$$=−\int\frac{{A}}{\mathrm{1}−{t}}{dt}\:−\int\frac{{B}}{\mathrm{1}+{t}}{dt}\:−\int\frac{{C}}{\mathrm{1}+\mathrm{2}{t}}{dt}= \\ $$$$={A}\:{ln}\left(\mathrm{1}−{t}\right)−{B}\:{ln}\left(\mathrm{1}+{t}\right)−\frac{{C}}{\mathrm{2}}\:{ln}\left(\mathrm{1}+\mathrm{2}{t}\right)+{c} \\ $$$${where} \\ $$$$\mathrm{1}={A}\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+\mathrm{2}{t}\right)+{B}\left(\mathrm{1}−{t}\right)\left(\mathrm{1}+\mathrm{2}{t}\right)+{C}\left(\mathrm{1}−{t}^{\mathrm{2}} \right) \\ $$$${and}\:{if}\:{t}=\mathrm{1}:\:\mathrm{1}={A}.\mathrm{2}.\mathrm{3}\Rightarrow{A}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${if}\:{t}=−\mathrm{1}:\:\mathrm{1}={B}\left(−\mathrm{2}\right)\left(−\mathrm{1}\right)\Rightarrow{B}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${if}\:{t}=−\frac{\mathrm{1}}{\mathrm{2}}:\:\mathrm{1}={C}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\Rightarrow{C}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${so} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\:{ln}\left(\mathrm{1}−{cos}\:{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{cos}\:{x}\right)−\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\mathrm{1}+\mathrm{2}{cos}\:{x}\right)+{c} \\ $$
Commented by sanusihammed last updated on 27/Jun/16
I really appreciate
$${I}\:{really}\:{appreciate} \\ $$

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