Question Number 53259 by Tawa1 last updated on 19/Jan/19
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jan/19
![I=∫_0 ^π ((xdx)/(a^2 cos^2 x+b^2 sin^2 x))dx I=∫_0 ^π (((π−x)dx)/(a^2 cos^2 (π−x)+b^2 sin^2 (π−x))) 2I=∫_0 ^π ((x+π−x)/(a^2 cos^2 x+b^2 sin^2 x))dx ((2I)/π)=∫_0 ^π (dx/(cos^2 x(a^2 +b^2 tan^2 x))) =∫_0 ^π ((sec^2 xdx)/(b^2 ((a^2 /b^2 )+tan^2 x))) ((2I)/π)=(1/b^2 )∫_0 ^π ((sec^2 x)/((a^2 /b^2 )+tan^2 x))dx wait... f(x)=((sec^2 x)/((a^2 /b^2 )+tan^2 x)) f(π−x)=((sec^2 (π−x))/((a^2 /b^2 )+tan^2 (π−x)))=f(x) so ((2I)/π)=(1/b^2 )×2∫_0 ^(π/2) ((sec^2 x)/((a^2 /b^2 )+tan^2 x)) ((Ib^2 )/π)=∫_0 ^(π/2) ((sec^2 x)/((a^2 /b^2 )+tan^2 x))dx let k=tanx [dk=sec^2 xdx ((Ib^2 )/π)=∫_( 0) ^∞ (dk/((a^2 /b^2 )+k^2 )) ((Ib^2 )/π)=(1/(a/b))×∣tan^(−1) ((k/(a/b)))∣_0 ^∞ I=(π/(ab))[tan^(−1) (∞)−tan^(−1) (0)] I=(π/(ab))×(π/2)=(π^2 /(2ab))](https://www.tinkutara.com/question/Q53268.png)
$${I}=\int_{\mathrm{0}} ^{\pi} \frac{{xdx}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{x}\right){dx}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\pi−{x}\right)+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} \left(\pi−{x}\right)} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\pi} \frac{{x}+\pi−{x}}{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\overset{\pi} {\int}_{\mathrm{0}} \frac{{dx}}{{cos}^{\mathrm{2}} {x}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} {tan}^{\mathrm{2}} {x}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {xdx}}{{b}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+{tan}^{\mathrm{2}} {x}\right)} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {x}}{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+{tan}^{\mathrm{2}} {x}}{dx} \\ $$$${wait}… \\ $$$${f}\left({x}\right)=\frac{{sec}^{\mathrm{2}} {x}}{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+{tan}^{\mathrm{2}} {x}} \\ $$$${f}\left(\pi−{x}\right)=\frac{{sec}^{\mathrm{2}} \left(\pi−{x}\right)}{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+{tan}^{\mathrm{2}} \left(\pi−{x}\right)}={f}\left({x}\right) \\ $$$${so} \\ $$$$\frac{\mathrm{2}{I}}{\pi}=\frac{\mathrm{1}}{{b}^{\mathrm{2}} }×\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} {x}}{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+{tan}^{\mathrm{2}} {x}} \\ $$$$\frac{{Ib}^{\mathrm{2}} }{\pi}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} {x}}{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+{tan}^{\mathrm{2}} {x}}{dx} \\ $$$${let}\:\:\:{k}={tanx}\:\:\left[{dk}={sec}^{\mathrm{2}} {xdx}\right. \\ $$$$\frac{{Ib}^{\mathrm{2}} }{\pi}=\int_{\:\mathrm{0}} ^{\infty} \frac{{dk}}{\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+{k}^{\mathrm{2}} } \\ $$$$\frac{{Ib}^{\mathrm{2}} }{\pi}=\frac{\mathrm{1}}{\frac{{a}}{{b}}}×\mid{tan}^{−\mathrm{1}} \left(\frac{{k}}{\frac{{a}}{{b}}}\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$${I}=\frac{\pi}{{ab}}\left[{tan}^{−\mathrm{1}} \left(\infty\right)−{tan}^{−\mathrm{1}} \left(\mathrm{0}\right)\right] \\ $$$${I}=\frac{\pi}{{ab}}×\frac{\pi}{\mathrm{2}}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}{ab}} \\ $$
Commented by Tawa1 last updated on 19/Jan/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$