Find-sin-x-3-cos-x-sin-3x-dx-

Question Number 163730 by HongKing last updated on 09/Jan/22

Find: 𝛀 = ∫ ((sin(x) + (√3) cos(x))/(sin(3x))) dx

$$\mathrm{Find}:\:\:\boldsymbol{\Omega}\:=\:\int\:\frac{\mathrm{sin}\left(\mathrm{x}\right)\:+\:\sqrt{\mathrm{3}}\:\mathrm{cos}\left(\mathrm{x}\right)}{\mathrm{sin}\left(\mathrm{3x}\right)}\:\mathrm{dx} \\ $$

Answered by cortano1 last updated on 10/Jan/22

=2∫ ((sin (x+(π/3)))/(sin (3x))) dx = 2∫ ((d(tan θ))/(3−tan^2 θ)) ; θ=x+(π/6) = ((2(√3))/3) tanh^(−1) (((tan θ)/( (√3)))) + c = ((2(√3))/3) tanh^(−1) (((tan (x+(π/6)))/( (√3)))) + c

$$\:=\mathrm{2}\int\:\frac{\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{3}}\right)}{\mathrm{sin}\:\left(\mathrm{3}{x}\right)}\:{dx} \\ $$$$\:=\:\mathrm{2}\int\:\frac{{d}\left(\mathrm{tan}\:\theta\right)}{\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} \theta}\:;\:\theta={x}+\frac{\pi}{\mathrm{6}} \\ $$$$\:=\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:\theta}{\:\sqrt{\mathrm{3}}}\right)\:+\:{c} \\ $$$$\:=\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\:\mathrm{tanh}\:^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{6}}\right)}{\:\sqrt{\mathrm{3}}}\right)\:+\:{c}\: \\ $$

Answered by MJS_new last updated on 10/Jan/22

∫((sin x +(√3)cos x)/(sin 3x))dx= =−∫((((√3)+tan x)(1+tan^2 x))/((3−tan x)tan x))dx= =∫((1+tan^2 x)/(((√3)−tan x)tan x))dx= [t=tan x → dx=(dt/(1+tan^2 x))] =∫(dt/(t((√3)−t)))=((√3)/3)ln ∣(t/(t−(√3)))∣ = =((√3)/3)ln ∣((sin x)/(sin x −(√3)cos x))∣ +C

$$\int\frac{\mathrm{sin}\:{x}\:+\sqrt{\mathrm{3}}\mathrm{cos}\:{x}}{\mathrm{sin}\:\mathrm{3}{x}}{dx}= \\ $$$$=−\int\frac{\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:{x}\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right)}{\left(\mathrm{3}−\mathrm{tan}\:{x}\right)\mathrm{tan}\:{x}}{dx}= \\ $$$$=\int\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}{\left(\sqrt{\mathrm{3}}−\mathrm{tan}\:{x}\right)\mathrm{tan}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}\right] \\ $$$$=\int\frac{{dt}}{{t}\left(\sqrt{\mathrm{3}}−{t}\right)}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{ln}\:\mid\frac{{t}}{{t}−\sqrt{\mathrm{3}}}\mid\:= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{ln}\:\mid\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}\:−\sqrt{\mathrm{3}}\mathrm{cos}\:{x}}\mid\:+{C} \\ $$

Commented by peter frank last updated on 11/Jan/22

$$\mathrm{great} \\ $$

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