M-dx-2cos-x-3sin-x-

Question Number 134660 by benjo_mathlover last updated on 06/Mar/21

$$\mathscr{M}\:=\:\int\:\frac{{dx}}{\mathrm{2cos}\:{x}+\mathrm{3sin}\:{x}}\: \\ $$

Answered by EDWIN88 last updated on 06/Mar/21

Let { ((2 = r sin α)),((3 = r cos α)) :} ⇒ r^2 = 13; r =(√(13)) and tan α = (2/3) ,∴ α = tan^(−1) ((2/3)) M =(1/r)∫ (dx/(sin (α+x))) = (1/r)∫ cosec (x+α) dx M = (1/r) ln [tan (((x+α))/2) ] + c M = (1/( (√(13)))) ln [ tan ((x/2)+(1/2)tan^(−1) ((2/3)))] + c

$$\mathrm{Let}\:\begin{cases}{\mathrm{2}\:=\:\mathrm{r}\:\mathrm{sin}\:\alpha}\\{\mathrm{3}\:=\:\mathrm{r}\:\mathrm{cos}\:\alpha}\end{cases}\:\Rightarrow\:\mathrm{r}^{\mathrm{2}} =\:\mathrm{13};\:\mathrm{r}\:=\sqrt{\mathrm{13}} \\ $$$$\:\mathrm{and}\:\mathrm{tan}\:\alpha\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:,\therefore\:\alpha\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\mathscr{M}\:=\frac{\mathrm{1}}{\mathrm{r}}\int\:\frac{\mathrm{dx}}{\mathrm{sin}\:\left(\alpha+\mathrm{x}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{r}}\int\:\mathrm{cosec}\:\left(\mathrm{x}+\alpha\right)\:\mathrm{dx} \\ $$$$\mathscr{M}\:=\:\frac{\mathrm{1}}{\mathrm{r}}\:\mathrm{ln}\:\left[\mathrm{tan}\:\frac{\left(\mathrm{x}+\alpha\right)}{\mathrm{2}}\:\right]\:+\:\mathrm{c} \\ $$$$\mathscr{M}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}}\:\mathrm{ln}\:\left[\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{3}}\right)\right)\right]\:+\:\mathrm{c} \\ $$

Answered by mathmax by abdo last updated on 07/Mar/21

M=∫ (dx/(2cosx+3sinx)) we do the changement tan((x/2))=t ⇒ M=∫ ((2dt)/((1+t^2 )(2((1−t^2 )/(1+t^2 ))+((6t)/(1+t^2 ))))) =∫ ((2dt)/(2(1−t^2 )+6t)) =∫ (dt/(1−t^2 +3t)) =−∫ (dt/(t^2 −3t−1)) Δ=9+4 =13 ⇒t_1 =((3+(√(13)))/2) and t_2 =((3−(√(13)))/2) ⇒ M=−∫ (dt/((t−t_1 )(t−t_2 ))) =−(1/(t_1 −t_2 ))∫ ((1/(t−t_1 ))−(1/(t−t_2 )))dt =−(1/( (√(13))))ln∣((t−t_1 )/(t−t_2 ))∣ +C =−(1/( (√(13))))ln∣((t−((3+(√(13)))/2))/(t−((3−(√(13)))/2)))∣ +C =−(1/( (√(13))))ln∣((2t−3−(√(13)))/(2t−3+(√(13))))∣ +C

$$\mathrm{M}=\int\:\frac{\mathrm{dx}}{\mathrm{2cosx}+\mathrm{3sinx}}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{M}=\int\:\:\frac{\mathrm{2dt}}{\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\left(\mathrm{2}\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }+\frac{\mathrm{6t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\right)}\:=\int\:\frac{\mathrm{2dt}}{\mathrm{2}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)+\mathrm{6t}}\:=\int\:\frac{\mathrm{dt}}{\mathrm{1}−\mathrm{t}^{\mathrm{2}} \:+\mathrm{3t}} \\ $$$$=−\int\:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{3t}−\mathrm{1}} \\ $$$$\Delta=\mathrm{9}+\mathrm{4}\:=\mathrm{13}\:\Rightarrow\mathrm{t}_{\mathrm{1}} =\frac{\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{t}_{\mathrm{2}} =\frac{\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{M}=−\int\:\:\frac{\mathrm{dt}}{\left(\mathrm{t}−\mathrm{t}_{\mathrm{1}} \right)\left(\mathrm{t}−\mathrm{t}_{\mathrm{2}} \right)}\:=−\frac{\mathrm{1}}{\mathrm{t}_{\mathrm{1}} −\mathrm{t}_{\mathrm{2}} }\int\:\left(\frac{\mathrm{1}}{\mathrm{t}−\mathrm{t}_{\mathrm{1}} }−\frac{\mathrm{1}}{\mathrm{t}−\mathrm{t}_{\mathrm{2}} }\right)\mathrm{dt} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}}\mathrm{ln}\mid\frac{\mathrm{t}−\mathrm{t}_{\mathrm{1}} }{\mathrm{t}−\mathrm{t}_{\mathrm{2}} }\mid\:+\mathrm{C}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}}\mathrm{ln}\mid\frac{\mathrm{t}−\frac{\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}}}{\mathrm{t}−\frac{\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}}\mid\:+\mathrm{C} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}}\mathrm{ln}\mid\frac{\mathrm{2t}−\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2t}−\mathrm{3}+\sqrt{\mathrm{13}}}\mid\:+\mathrm{C} \\ $$

Commented by mathmax by abdo last updated on 07/Mar/21

t=tan((x/2)) ⇒M =−(1/( (√(13))))ln∣((2tan((x/2))−3−(√(13)))/(2tan((x/2))−3+(√(13))))∣ +C

$$\mathrm{t}=\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:\Rightarrow\mathrm{M}\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}}\mathrm{ln}\mid\frac{\mathrm{2tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{3}+\sqrt{\mathrm{13}}}\mid\:+\mathrm{C} \\ $$

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