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Question Number 200318 by sonukgindia last updated on 17/Nov/23

Answered by Sutrisno last updated on 17/Nov/23

=∫_0 ^π ((1/(cos^2 x))/((1/(cos^2 x))+((cos^2 x)/(cos^2 x))))dx =∫_0 ^π ((sec^2 x)/(tan^2 x+2))dx misal=:tanx=(√2)tanθ→dx=(((√2)sec^2 θ)/(sec^2 x))dθ =∫((sec^2 x)/(((√2)tanθ)^2 +2)).(((√2)sec^2 θ)/(sec^2 x))dθ =∫(((√2)sec^2 θ)/(2sec^2 θ))dθ =((√2)/2)∫dθ =((√2)/2)θ =((√2)/2)arctan(((tanx)/( (√2))))∣_0 ^π =0

$$=\int_{\mathrm{0}} ^{\pi} \frac{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}}{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}+\frac{{cos}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {x}}{{tan}^{\mathrm{2}} {x}+\mathrm{2}}{dx} \\ $$$${misal}=:{tanx}=\sqrt{\mathrm{2}}{tan}\theta\rightarrow{dx}=\frac{\sqrt{\mathrm{2}}{sec}^{\mathrm{2}} \theta}{{sec}^{\mathrm{2}} {x}}{d}\theta \\ $$$$=\int\frac{{sec}^{\mathrm{2}} {x}}{\left(\sqrt{\mathrm{2}}{tan}\theta\right)^{\mathrm{2}} +\mathrm{2}}.\frac{\sqrt{\mathrm{2}}{sec}^{\mathrm{2}} \theta}{{sec}^{\mathrm{2}} {x}}{d}\theta \\ $$$$=\int\frac{\sqrt{\mathrm{2}}{sec}^{\mathrm{2}} \theta}{\mathrm{2}{sec}^{\mathrm{2}} \theta}{d}\theta \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int{d}\theta \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\theta \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{arctan}\left(\frac{{tanx}}{\:\sqrt{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$$=\mathrm{0} \\ $$

Commented by mr W last updated on 17/Nov/23

(1/2)<(1/(1+cos^2 x))<1 ⇒(π/2)<∫_0 ^π (dx/(1+cos^2 x))<π that means ∫_0 ^π (dx/(1+cos^2 x)) can never be 0.

$$\frac{\mathrm{1}}{\mathrm{2}}<\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}<\mathrm{1} \\ $$$$\Rightarrow\frac{\pi}{\mathrm{2}}<\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}<\pi \\ $$$${that}\:{means}\:\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}\:{can}\:{never}\:{be}\:\mathrm{0}. \\ $$

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 ✓

$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{tan}^{\mathrm{2}} {x}}{\mathrm{2}+{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$\left.{tanx}={u}\Rightarrow{I}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\frac{{du}}{\mathrm{2}+{u}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\pi\:\checkmark \\ $$

Commented by MM42 last updated on 17/Nov/23

★★★ Attention ★★★ I_7 =∫_0 ^π (dx/(1+cos^2 x)) =2∫_0 ^(π/2) (dx/(1+cos^2 x)) =2∫_0 ^(π/2) (dx/(1+sin^2 ((π/2)−x))) (π/2)−x =u⇒I_7 =2∫_0 ^(π/2) (dx/(1+sin^2 x)) =I_8

$$\bigstar\bigstar\bigstar\:\:\:\:{Attention}\:\:\:\:\:\:\bigstar\bigstar\bigstar \\ $$$${I}_{\mathrm{7}} =\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}}\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}\: \\ $$$$\frac{\pi}{\mathrm{2}}−{x}\:={u}\Rightarrow{I}_{\mathrm{7}} =\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}\:={I}_{\mathrm{8}} \: \\ $$$$\: \\ $$

Commented by MM42 last updated on 17/Nov/23

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