Question Number 179042 by mnjuly1970 last updated on 23/Oct/22
Answered by Ar Brandon last updated on 23/Oct/22
![I=∫_0 ^(π/2) ((cos^2 x)/(sin^4 x+cos^2 x))dx=∫_0 ^∞ (dt/(t^4 +t^2 +1)) =(1/2)∫_0 ^∞ ((1+(1/t^2 ))/((t−(1/t))^2 +3))dt−(1/2)∫_0 ^∞ ((1−(1/t^2 ))/((t+(1/t))^2 −1))dt =(1/(2(√3)))[arctan(((t^2 −1)/(t(√3))))]_0 ^∞ +(1/4)[ln∣((t^2 +t+1)/(t^2 −t+1))∣]_0 ^∞ =(1/(2(√3)))((π/2)−−(π/2))=(π/(2(√3))) ⇒a=((2(√3))/π)](Q179045.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{cos}^{\mathrm{2}} {x}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{3}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left[\mathrm{arctan}\left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{ln}\mid\frac{{t}^{\mathrm{2}} +{t}+\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{\infty} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left(\frac{\pi}{\mathrm{2}}−−\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:\Rightarrow{a}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\pi} \\ $$
Commented by mnjuly1970 last updated on 24/Oct/22
$$\:{thank}\:{you}\:{so}\:{much}\:{sir}\:{Brandon} \\ $$
Answered by MJS_new last updated on 23/Oct/22
![∫((cos^2 x)/(sin^4 x +cos^2 x))dx= [t=tan x → dx=cos^2 x dt] =∫_0 ^(π/2) (dt/(t^4 +t^2 +1))= =(1/4)∫_0 ^∞ (((2t+1)/(t^2 +t+1))+(1/(t^2 +t+1))−((2t−1)/(t^2 −t+1))+(1/(t^2 −t+1)))dt= =(1/4)[ln (t^2 +t+1) +((2(√3))/3)arctan (((√3)(2t+1))/3) −ln (t^2 −t+1) +((2(√3))/3)arctan (((√3)(2t−1))/3)]_0 ^∞ = =(((√3)π)/6) ⇒ a=(6/( (√3)π))=((2(√3))/π)](Q179046.png)
$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}\right] \\ $$$$=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{dt}}{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{2}{t}+\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}−\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} −{t}+\mathrm{1}}\right){dt}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{ln}\:\left({t}^{\mathrm{2}} +{t}+\mathrm{1}\right)\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}+\mathrm{1}\right)}{\mathrm{3}}\:−\mathrm{ln}\:\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)\:+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\:\frac{\sqrt{\mathrm{3}}\left(\mathrm{2}{t}−\mathrm{1}\right)}{\mathrm{3}}\right]_{\mathrm{0}} ^{\infty} = \\ $$$$=\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{6}} \\ $$$$\Rightarrow \\ $$$${a}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{3}}\pi}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\pi} \\ $$
Commented by mnjuly1970 last updated on 24/Oct/22
$${grateful}\:{sir}\:{very}\:{nice}\:{solution} \\ $$
Commented by SLVR last updated on 24/Oct/22
$${great}\:{sir} \\ $$