Question Number 29445 by prof Abdo imad last updated on 08/Feb/18
$${find}\:\:\:\int\:\:\:\:\:\:\:\frac{{dx}}{{sinx}\:+{sin}\left(\mathrm{2}{x}\right)}\:. \\ $$
Answered by mrW2 last updated on 09/Feb/18
![=∫(dx/(sin x(1+2cos x))) =∫((sin x dx)/(sin^2 x(1+2cos x))) =∫((d(cos x))/((cos^2 x−1)(2cos x+1))) =∫(dt/((t^2 −1)(2t+1))) =∫[−(1/(2(t+1)))−(1/(6(t−1)))+(4/(3(2t+1)))]dt =−((ln (t+1))/2)−((ln (t−1))/6)+((2ln (2t+1))/3)+C =−((3ln (t+1))/6)−((ln (t−1))/6)+((4ln (2t+1))/6)+C =−((ln (t+1)^3 )/6)−((ln (t−1))/6)+((ln (2t+1)^4 )/6)+C =(1/6)ln∣ (((2t+1)^4 )/((t+1)^3 (t−1)))∣+C =(1/6)ln∣ (((2cos x+1)^4 )/((cos x+1)^3 (cos x−1)))∣+C =(1/6)ln∣ (((2cos x+1)^4 )/((cos x+1)^2 (sin^2 x)))∣+C =(1/3)ln∣ (((2cos x+1)^2 )/(sin x (cos x+1)))∣+C](https://www.tinkutara.com/question/Q29540.png)
$$=\int\frac{{dx}}{\mathrm{sin}\:{x}\left(\mathrm{1}+\mathrm{2cos}\:{x}\right)} \\ $$$$=\int\frac{\mathrm{sin}\:{x}\:{dx}}{\mathrm{sin}^{\mathrm{2}} \:{x}\left(\mathrm{1}+\mathrm{2cos}\:{x}\right)} \\ $$$$=\int\frac{{d}\left(\mathrm{cos}\:{x}\right)}{\left(\mathrm{cos}^{\mathrm{2}} \:{x}−\mathrm{1}\right)\left(\mathrm{2cos}\:{x}+\mathrm{1}\right)} \\ $$$$=\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)} \\ $$$$=\int\left[−\frac{\mathrm{1}}{\mathrm{2}\left({t}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{6}\left({t}−\mathrm{1}\right)}+\frac{\mathrm{4}}{\mathrm{3}\left(\mathrm{2}{t}+\mathrm{1}\right)}\right]{dt} \\ $$$$=−\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{\mathrm{2}}−\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{\mathrm{6}}+\frac{\mathrm{2ln}\:\left(\mathrm{2}{t}+\mathrm{1}\right)}{\mathrm{3}}+{C} \\ $$$$=−\frac{\mathrm{3ln}\:\left({t}+\mathrm{1}\right)}{\mathrm{6}}−\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{\mathrm{6}}+\frac{\mathrm{4ln}\:\left(\mathrm{2}{t}+\mathrm{1}\right)}{\mathrm{6}}+{C} \\ $$$$=−\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{6}}−\frac{\mathrm{ln}\:\left({t}−\mathrm{1}\right)}{\mathrm{6}}+\frac{\mathrm{ln}\:\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{4}} }{\mathrm{6}}+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\mid\:\frac{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{4}} }{\left({t}+\mathrm{1}\right)^{\mathrm{3}} \left({t}−\mathrm{1}\right)}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\mid\:\frac{\left(\mathrm{2cos}\:{x}+\mathrm{1}\right)^{\mathrm{4}} }{\left(\mathrm{cos}\:{x}+\mathrm{1}\right)^{\mathrm{3}} \left(\mathrm{cos}\:{x}−\mathrm{1}\right)}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{6}}\mathrm{ln}\mid\:\frac{\left(\mathrm{2cos}\:{x}+\mathrm{1}\right)^{\mathrm{4}} }{\left(\mathrm{cos}\:{x}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:{x}\right)}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\mid\:\frac{\left(\mathrm{2cos}\:{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{sin}\:{x}\:\left(\mathrm{cos}\:{x}+\mathrm{1}\right)}\mid+{C} \\ $$