Question Number 68967 by aliesam last updated on 17/Sep/19
Answered by mind is power last updated on 17/Sep/19
![tg((x/2))=t dx=((2dt)/(1+t^2 )) ⇒sin(x)=((2t)/(1+t^2 )) cos(x)=((1−t^2 )/(1+t^2 )) ⇒∫(dx/(5sin(x)−4cos(x)+3))=∫((2dt)/(3(1+t^2 )+10t−4+4t^2 )) =∫(dt/(7t^2 +10t−1))=∫(dt/(7(t+5−(√(32)))(t+5+(√(32))))) =∫[(1/(14(√(32))(t+5−(√(32)))))−(1/(14(√(32))(t+5+(√(32)))))]dt =(1/(14(√(32))))ln∣((t+5−(√(32)))/(t+5+(√(32))))∣+c ⇒∫(dx/(5sin(x)−4cos(x)+3))=(1/(14(√(32))))ln∣((tan((x/2))+5−(√(32)))/(tg((x/2))+5+(√(32))))∣+c](https://www.tinkutara.com/question/Q69017.png)
$${tg}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\ $$$${dx}=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow{sin}\left({x}\right)=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:{cos}\left({x}\right)=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\int\frac{{dx}}{\mathrm{5}{sin}\left({x}\right)−\mathrm{4}{cos}\left({x}\right)+\mathrm{3}}=\int\frac{\mathrm{2}{dt}}{\mathrm{3}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+\mathrm{10}{t}−\mathrm{4}+\mathrm{4}{t}^{\mathrm{2}} } \\ $$$$=\int\frac{{dt}}{\mathrm{7}{t}^{\mathrm{2}} +\mathrm{10}{t}−\mathrm{1}}=\int\frac{{dt}}{\mathrm{7}\left({t}+\mathrm{5}−\sqrt{\mathrm{32}}\right)\left({t}+\mathrm{5}+\sqrt{\mathrm{32}}\right)} \\ $$$$=\int\left[\frac{\mathrm{1}}{\mathrm{14}\sqrt{\mathrm{32}}\left({t}+\mathrm{5}−\sqrt{\mathrm{32}}\right)}−\frac{\mathrm{1}}{\mathrm{14}\sqrt{\mathrm{32}}\left({t}+\mathrm{5}+\sqrt{\left.\mathrm{32}\right)}\right.}\right]{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{14}\sqrt{\mathrm{32}}}{ln}\mid\frac{{t}+\mathrm{5}−\sqrt{\mathrm{32}}}{{t}+\mathrm{5}+\sqrt{\mathrm{32}}}\mid+{c} \\ $$$$ \\ $$$$\Rightarrow\int\frac{{dx}}{\mathrm{5}{sin}\left({x}\right)−\mathrm{4}{cos}\left({x}\right)+\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{14}\sqrt{\mathrm{32}}}{ln}\mid\frac{{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{5}−\sqrt{\mathrm{32}}}{{tg}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{5}+\sqrt{\mathrm{32}}}\mid+{c} \\ $$$$ \\ $$