Question Number 160767 by cortano last updated on 06/Dec/21
$$\:\:\:\:\int_{\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=?\: \\ $$
Answered by MJS_new last updated on 06/Dec/21
![∫_0 ^(π/2) ((cos^2 x)/(cos^2 x +4sin^2 x))dx= [t=tan x → dx=(dt/(t^2 +1))] =∫_0 ^∞ (dt/((t^2 +1)(4t^2 +1)))=(1/3)∫_0 ^∞ ((1/(t^2 +(1/4)))−(1/(t^2 +1)))dt= =(1/3)[2arctan 2t −arctan t]_0 ^∞ =(π/6)](Q160778.png)
$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{cos}^{\mathrm{2}} \:{x}\:+\mathrm{4sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\ $$$$=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[\mathrm{2arctan}\:\mathrm{2}{t}\:−\mathrm{arctan}\:{t}\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mathrm{6}} \\ $$