Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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+(Ο€/3)))/(sin (3x))) dx = 2∫ ((d(tan ΞΈ))/(3βˆ’tan^2 ΞΈ)) ; ΞΈ=x+(Ο€/6) = ((2(√3))/3) tanh^(βˆ’1) (((tan ΞΈ)/( (√3)))) + c = ((2(√3))/3) tanh^(βˆ’1) (((tan (x+(Ο€/6)))/( (√3)))) + c

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

Answered by MJS_new last updated on 10/Jan/22

∫((sin x +(√3)cos x)/(sin 3x))dx= =βˆ’βˆ«((((√3)+tan x)(1+tan^2 x))/((3βˆ’tan x)tan x))dx= =∫((1+tan^2 x)/(((√3)βˆ’tan x)tan x))dx= [t=tan x β†’ dx=(dt/(1+tan^2 x))] =∫(dt/(t((√3)βˆ’t)))=((√3)/3)ln ∣(t/(tβˆ’(√3)))∣ = =((√3)/3)ln ∣((sin x)/(sin x βˆ’(√3)cos x))∣ +C

$$\int\frac{\mathrm{sin}\:{x}\:+\sqrt{\mathrm{3}}\mathrm{cos}\:{x}}{\mathrm{sin}\:\mathrm{3}{x}}{dx}= \\ $$$$=βˆ’\int\frac{\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:{x}\right)\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right)}{\left(\mathrm{3}βˆ’\mathrm{tan}\:{x}\right)\mathrm{tan}\:{x}}{dx}= \\ $$$$=\int\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}{\left(\sqrt{\mathrm{3}}βˆ’\mathrm{tan}\:{x}\right)\mathrm{tan}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}}\right] \\ $$$$=\int\frac{{dt}}{{t}\left(\sqrt{\mathrm{3}}βˆ’{t}\right)}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{ln}\:\mid\frac{{t}}{{t}βˆ’\sqrt{\mathrm{3}}}\mid\:= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{ln}\:\mid\frac{\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}\:βˆ’\sqrt{\mathrm{3}}\mathrm{cos}\:{x}}\mid\:+{C} \\ $$

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

great

$$\mathrm{great} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com