Question Number 200318 by sonukgindia last updated on 17/Nov/23
Answered by Sutrisno last updated on 17/Nov/23
$$=\int_{\mathrm{0}} ^{\pi} \frac{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}}{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}+\frac{{cos}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {x}}{{tan}^{\mathrm{2}} {x}+\mathrm{2}}{dx} \\ $$$${misal}=:{tanx}=\sqrt{\mathrm{2}}{tan}\theta\rightarrow{dx}=\frac{\sqrt{\mathrm{2}}{sec}^{\mathrm{2}} \theta}{{sec}^{\mathrm{2}} {x}}{d}\theta \\ $$$$=\int\frac{{sec}^{\mathrm{2}} {x}}{\left(\sqrt{\mathrm{2}}{tan}\theta\right)^{\mathrm{2}} +\mathrm{2}}.\frac{\sqrt{\mathrm{2}}{sec}^{\mathrm{2}} \theta}{{sec}^{\mathrm{2}} {x}}{d}\theta \\ $$$$=\int\frac{\sqrt{\mathrm{2}}{sec}^{\mathrm{2}} \theta}{\mathrm{2}{sec}^{\mathrm{2}} \theta}{d}\theta \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int{d}\theta \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\theta \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{arctan}\left(\frac{{tanx}}{\:\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$$=\mathrm{0} \\ $$
Commented by mr W last updated on 17/Nov/23
$$\frac{\mathrm{1}}{\mathrm{2}}<\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}<\mathrm{1} \\ $$$$\Rightarrow\frac{\pi}{\mathrm{2}}<\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}<\pi \\ $$$${that}\:{means}\:\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}\:{can}\:{never}\:{be}\:\mathrm{0}. \\ $$
Answered by MM42 last updated on 17/Nov/23
![I=2∫_0 ^(π/2) (dx/(1+cos^2 x)) =2∫_0 ^(π/2) ((1+tan^2 x)/(2+tan^2 x))dx tanx=u⇒I=2∫_0 ^∞ (du/(2+u^2 )) =(√2)tan^(−1) ((u/( (√2))))]_0 ^∞ =((√2)/2) π ✓](https://www.tinkutara.com/question/Q200343.png)
$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {x}}{\mathrm{2}+{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\left.{tanx}={u}\Rightarrow{I}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\frac{{du}}{\mathrm{2}+{u}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\pi\:\checkmark \\ $$
Commented by MM42 last updated on 17/Nov/23
$$\bigstar\bigstar\bigstar\:\:\:\:{Attention}\:\:\:\:\:\:\bigstar\bigstar\bigstar \\ $$$${I}_{\mathrm{7}} =\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}\: \\ $$$$\frac{\pi}{\mathrm{2}}−{x}\:={u}\Rightarrow{I}_{\mathrm{7}} =\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}\:={I}_{\mathrm{8}} \: \\ $$$$\: \\ $$
Commented by MM42 last updated on 17/Nov/23
