Question Number 221601 by Nicholas666 last updated on 08/Jun/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{{sin}\:\mathrm{2}{x}}{\mathrm{1}\:+\:{sin}\:\mathrm{3}{x}}\:{dx} \\ $$$$ \\ $$
Answered by Frix last updated on 08/Jun/25
![=−2∫((cos x sin x)/(4sin^3 x −3sin x +1))dx =^([t=sin x]) =−2∫(t/((t−1)(2t+1)^2 ))dt =^([u=((t−1)/(2t+1))]) =−(2/9)∫(1+(1/u))du=−((2u)/9)−(2/9)ln u = =(2/9)ln ((2t+1)/(t−1)) −((2(t−1))/(9(2t+1)))= =((2(1−sin x))/(9(1+2sin x)))+(2/9)ln ∣((1+2sin x)/(1−sin x))∣ +C](https://www.tinkutara.com/question/Q221613.png)
$$=−\mathrm{2}\int\frac{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}{\mathrm{4sin}^{\mathrm{3}} \:{x}\:−\mathrm{3sin}\:{x}\:+\mathrm{1}}{dx}\:\overset{\left[{t}=\mathrm{sin}\:{x}\right]} {=} \\ $$$$=−\mathrm{2}\int\frac{{t}}{\left({t}−\mathrm{1}\right)\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\overset{\left[{u}=\frac{{t}−\mathrm{1}}{\mathrm{2}{t}+\mathrm{1}}\right]} {=} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{9}}\int\left(\mathrm{1}+\frac{\mathrm{1}}{{u}}\right){du}=−\frac{\mathrm{2}{u}}{\mathrm{9}}−\frac{\mathrm{2}}{\mathrm{9}}\mathrm{ln}\:{u}\:= \\ $$$$=\frac{\mathrm{2}}{\mathrm{9}}\mathrm{ln}\:\frac{\mathrm{2}{t}+\mathrm{1}}{{t}−\mathrm{1}}\:−\frac{\mathrm{2}\left({t}−\mathrm{1}\right)}{\mathrm{9}\left(\mathrm{2}{t}+\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{sin}\:{x}\right)}{\mathrm{9}\left(\mathrm{1}+\mathrm{2sin}\:{x}\right)}+\frac{\mathrm{2}}{\mathrm{9}}\mathrm{ln}\:\mid\frac{\mathrm{1}+\mathrm{2sin}\:{x}}{\mathrm{1}−\mathrm{sin}\:{x}}\mid\:+{C} \\ $$
Answered by mehdee7396 last updated on 08/Jun/25
$$\int\frac{\mathrm{2}{sinxcosx}}{\mathrm{1}+\mathrm{3}{sinx}−\mathrm{4}{sin}^{\mathrm{3}} {x}}{dx}\:\:\:\:\:{sinx}={u} \\ $$$$\Rightarrow\int\frac{\mathrm{2}{udu}}{\mathrm{1}+\mathrm{3}{u}−\mathrm{4}{u}^{\mathrm{3}} }=\int\frac{\mathrm{2}{udu}}{\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+\mathrm{2}{u}\right)^{\mathrm{2}} } \\ $$$$…\:{has}\:{a}\:{simple}\:{solution} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$