Question Number 177296 by peter frank last updated on 03/Oct/22
$$\:\:\mathrm{Evaluate}\: \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{x}}{\mathrm{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \mathrm{x}+\mathrm{b}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\mathrm{dx} \\ $$$$ \\ $$
Commented by peter frank last updated on 03/Oct/22
$$\mathrm{ans}\:\:\:\:\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{2ab}} \\ $$
Answered by Ar Brandon last updated on 03/Oct/22
![I=∫_0 ^π (x/(a^2 cos^2 x+b^2 sin^2 x))dx=∫_0 ^π ((π−x)/(a^2 cos^2 x+b^2 sin^2 x))dx =(π/2)∫_0 ^π (1/(a^2 cos^2 x+b^2 sin^2 x))dx=(π/2)∫_0 ^π ((sec^2 x)/(a^2 +b^2 tan^2 x))dx =π∫_0 ^(π/2) ((d(tanx))/(a^2 +b^2 tan^2 x))=(π/(ab))[arctan(((btanx)/a))]_0 ^(π/2) =(π^2 /(2ab))](https://www.tinkutara.com/question/Q177300.png)
$${I}=\int_{\mathrm{0}} ^{\pi} \frac{{x}}{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}{dx}=\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\:\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{{a}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} {x}}{dx}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{sec}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\:\:=\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\mathrm{tan}{x}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} {x}}=\frac{\pi}{{ab}}\left[\mathrm{arctan}\left(\frac{{b}\mathrm{tan}{x}}{{a}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$
Commented by peter frank last updated on 03/Oct/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by BaliramKumar last updated on 03/Oct/22
![I = ∫_0 ^π (x/(a^2 cos^2 x+b^2 sin^2 x))dx ..................[i] I = ∫_0 ^π (((π−x))/(a^2 cos^2 (π−x)+b^2 sin^2 (π−x)))dx I = ∫_0 ^π ((π−x)/(a^2 cos^2 x+b^2 sin^2 x))dx I = ∫_0 ^π (π/(a^2 cos^2 x+b^2 sin^2 x))dx− ∫_0 ^π (x/(a^2 cos^2 x+b^2 sin^2 x))dx I = ∫_0 ^π (π/(a^2 cos^2 x+b^2 sin^2 x))dx− I 2I = π∫_0 ^π (1/(a^2 cos^2 x+b^2 sin^2 x))dx I = (π/2)∫_0 ^π ((sec^2 x)/(a^2 +b^2 tan^2 x))dx I = (π/2)∫_0 ^π ((sec^2 (π−x))/(a^2 +b^2 tan^2 (π−x)))dx I = (π/2)∙2∫_0 ^(π/2) ((sec^2 x)/(a^2 +b^2 tan^2 x))dx I = π∫_0 ^(π/2) ((sec^2 x)/(a^2 + (btanx)^2 ))dx let b∙tan(x) = y when x=0 then y=0 sec^2 x∙dx = (dy/b) when x=(π/2) then y=∞ I = (π/b)∫_0 ^∞ (1/(a^2 +y^2 ))dy I = (π/b)∙(1/a)[tan^(−1) ((y/a))]_0 ^∞ I = (π/(ab))[(π/2)−0] I = (π^2 /(2ab))](https://www.tinkutara.com/question/Q177302.png)
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{{x}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx}\:\:\:………………\left[{i}\right] \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right)}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\pi−{x}\right)+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \left(\pi−{x}\right)}{dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{\pi}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx}−\:\int_{\mathrm{0}} ^{\pi} \frac{{x}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\:\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\pi} \frac{\pi}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx}−\:\mathrm{I} \\ $$$$\:\mathrm{2I}\:=\:\pi\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx} \\ $$$${I}\:=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {tan}^{\mathrm{2}} {x}}{dx} \\ $$$${I}\:=\:\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} \left(\pi−{x}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {tan}^{\mathrm{2}} \left(\pi−{x}\right)}{dx} \\ $$$${I}\:=\:\frac{\pi}{\mathrm{2}}\centerdot\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} {tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\mathrm{I}\:=\:\pi\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} {x}}{{a}^{\mathrm{2}} \:+\:\left(\mathrm{b}{tanx}\right)^{\mathrm{2}} }{dx} \\ $$$${let}\:\:\:\:\:\:\:\:{b}\centerdot{tan}\left({x}\right)\:=\:{y}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{when}\:\:{x}=\mathrm{0}\:\:{then}\:{y}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{sec}^{\mathrm{2}} {x}\centerdot{dx}\:=\:\frac{{dy}}{{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:{when}\:\:{x}=\frac{\pi}{\mathrm{2}}\:\:\:{then}\:\:\:{y}=\infty \\ $$$$\mathrm{I}\:=\:\frac{\pi}{{b}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{a}^{\mathrm{2}} +{y}^{\mathrm{2}} }{dy} \\ $$$$\mathrm{I}\:=\:\frac{\pi}{{b}}\centerdot\frac{\mathrm{1}}{{a}}\left[{tan}^{−\mathrm{1}} \left(\frac{{y}}{{a}}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$${I}\:=\:\frac{\pi}{{ab}}\left[\frac{\pi}{\mathrm{2}}−\mathrm{0}\right] \\ $$$${I}\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$
Commented by peter frank last updated on 03/Oct/22
$$\mathrm{thank}\:\mathrm{you} \\ $$