Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 141533 by sarkor last updated on 20/May/21

Answered by qaz last updated on 20/May/21

∫(dx/(1+3cos^2 x))=∫((sin^2 x+cos^2 x)/(sin^2 x+4cos^2 x))dx =∫((tan^2 x+1)/(tan^2 x+4))dx=∫((d(tan x))/(tan^2 x+4))=(1/2)tan^(−1) ((tan x)/2)+C

$$\int\frac{{dx}}{\mathrm{1}+\mathrm{3cos}\:^{\mathrm{2}} {x}}=\int\frac{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{4cos}\:^{\mathrm{2}} {x}}{dx} \\ $$$$=\int\frac{\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{4}}{dx}=\int\frac{{d}\left(\mathrm{tan}\:{x}\right)}{\mathrm{tan}\:^{\mathrm{2}} {x}+\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{tan}\:{x}}{\mathrm{2}}+{C} \\ $$

Commented by Mathspace last updated on 20/May/21

Ψ=∫ (dx/(1+3cos^2 x)) ⇒ Ψ=∫ (dx/(1+(3/2)(1+cos(2x)))) =_(2x=t) ∫ (dt/(2(1+(3/2)(1+cost)))) =∫ (dt/(2+3+3cost)) =∫ (dt/(5+3cost)) =_(tan((t/2))=y) ∫ ((2dy)/((1+y^2 )(5+3((1−y^2 )/(1+y^2 ))))) =∫ ((2dy)/(5+5y^2 +3−3y^2 )) =∫ ((2dy)/(8+2y^2 )) =∫ (dy/(4+y^2 )) =_(y=2z) =∫ ((2dz)/(4(1+z^2 ))) =(1/2)arctan(z) +C =(1/2)arctan((y/2)) +C =(1/2)arctan((1/2)tan(x)) +C

$$\Psi=\int\:\:\frac{{dx}}{\mathrm{1}+\mathrm{3}{cos}^{\mathrm{2}} {x}}\:\Rightarrow \\ $$$$\Psi=\int\:\:\frac{{dx}}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right)} \\ $$$$=_{\mathrm{2}{x}={t}} \:\:\int\:\:\frac{{dt}}{\mathrm{2}\left(\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}+{cost}\right)\right)} \\ $$$$=\int\:\:\frac{{dt}}{\mathrm{2}+\mathrm{3}+\mathrm{3}{cost}}\:=\int\:\frac{{dt}}{\mathrm{5}+\mathrm{3}{cost}} \\ $$$$=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={y}} \:\:\int\:\:\frac{\mathrm{2}{dy}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{5}+\mathrm{3}\frac{\mathrm{1}−{y}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }\right)} \\ $$$$=\int\:\:\frac{\mathrm{2}{dy}}{\mathrm{5}+\mathrm{5}{y}^{\mathrm{2}} \:+\mathrm{3}−\mathrm{3}{y}^{\mathrm{2}} } \\ $$$$=\int\:\:\frac{\mathrm{2}{dy}}{\mathrm{8}+\mathrm{2}{y}^{\mathrm{2}} }\:=\int\:\frac{{dy}}{\mathrm{4}+{y}^{\mathrm{2}} }\:=_{{y}=\mathrm{2}{z}} \\ $$$$=\int\:\frac{\mathrm{2}{dz}}{\mathrm{4}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left({z}\right)\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{{y}}{\mathrm{2}}\right)\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}{tan}\left({x}\right)\right)\:+{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com