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Question-200319

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Question Number 200319 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))+((sin^2 x)/(cos^2 x))))dx =∫_0 ^π ((sec^2 x)/(sec^2 x+tan^2 x))dx =∫_0 ^π ((sec^2 x)/(1+2tan^2 x))dx misal : tanθ=(√2)tanx→((sec^2 θ)/( (√2)sec^2 x))dθ=dx =∫((sec^2 x)/(1+2(((tanθ)/( (√2))))^2 )).((sec^2 θ)/( (√2)sec^2 x))dθ =(1/( (√2)))∫((sec^2 θ)/(1+tan^2 θ))dθ =(1/( (√2)))∫dθ =(1/( (√2)))θ =(1/( (√2)))arctan((√2)tanx)∣_0 ^π =(1/( (√2)))[arctan((√2)tanπ)−arctan((√2)tan0)] =0
$$=\int_{\mathrm{0}} ^{\pi} \frac{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}}{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}+\frac{{sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{2}} {x}}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {x}}{{sec}^{\mathrm{2}} {x}+{tan}^{\mathrm{2}} {x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{sec}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{2}{tan}^{\mathrm{2}} {x}}{dx} \\ $$$${misal}\::\:{tan}\theta=\sqrt{\mathrm{2}}{tanx}\rightarrow\frac{{sec}^{\mathrm{2}} \theta}{\:\sqrt{\mathrm{2}}{sec}^{\mathrm{2}} {x}}{d}\theta={dx} \\ $$$$=\int\frac{{sec}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{2}\left(\frac{{tan}\theta}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }.\frac{{sec}^{\mathrm{2}} \theta}{\:\sqrt{\mathrm{2}}{sec}^{\mathrm{2}} {x}}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{{sec}^{\mathrm{2}} \theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int{d}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\theta \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{arctan}\left(\sqrt{\mathrm{2}}{tanx}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[{arctan}\left(\sqrt{\mathrm{2}}{tan}\pi\right)−{arctan}\left(\sqrt{\mathrm{2}}{tan}\mathrm{0}\right)\right] \\ $$$$=\mathrm{0} \\ $$$$ \\ $$
Commented by mr W last updated on 17/Nov/23
(1/2)<(1/(1+sin^2 x))<1 ⇒(π/2)<∫_0 ^π (dx/(1+sin^2 x))<π that means ∫_0 ^π (dx/(1+sin^2 x)) can never be 0.
$$\frac{\mathrm{1}}{\mathrm{2}}<\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}<\mathrm{1} \\ $$$$\Rightarrow\frac{\pi}{\mathrm{2}}<\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}<\pi \\ $$$${that}\:{means}\:\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}\:{can}\:{never}\:{be}\:\mathrm{0}. \\ $$
Answered by MM42 last updated on 17/Nov/23
I=∫_0 ^π (1/(1+sin^2 x))dx=2∫_0 ^(π/2) (1/(1+sin^2 x))dx =2∫_0 ^(π/2) ((1+cot^2 x)/(2+cot^2 x))dx ; cotx=u ⇒I=2∫_0 ^∞ (du/(2+u^2 )) du=(√2) tan^(−1) ((u/( (√2))))]_0 ^∞ =((√2)/2)π ✓
$${I}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{cot}^{\mathrm{2}} {x}}{\mathrm{2}+{cot}^{\mathrm{2}} {x}}{dx}\:\:\:\:\:;\:\:\:{cotx}={u} \\ $$$$\left.\Rightarrow{I}=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{{du}}{\mathrm{2}+{u}^{\mathrm{2}} }\:{du}=\sqrt{\mathrm{2}}\:{tan}^{−\mathrm{1}} \left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\pi\:\:\checkmark \\ $$$$ \\ $$
Answered by Mathspace last updated on 18/Nov/23
I=∫_0 ^π (dx/(1+sin^2 x))=∫_0 ^π (dx/(1+((1−cos(2x))/2))) =2∫_0 ^π (dx/(3−cos(2x))) (2x=t) =∫_0 ^(2π) (dt/(3−cost))=∫_0 ^(2π) (dt/(3−((e^(it) +e^(−it) )/2))) =∫_0 ^(2π) ((2dt)/(6−e^(it) −e^(−it ) )) (e^(it) =z) =∫_(∣z∣=1) ((2dz)/(iz(6−z−z^(−1) ))) =∫_(∣z∣=1) ((−2i dz)/(6z−z^2 −1)) =∫_(∣z∣=1) ((2idz)/(z^2 −6z +1)) f(z)=((2i)/(z^2 −6z+1)) les poles? Δ^′ =3^2 −1=8 ⇒ z_1 =3+(√8)=3+2(√2) z_2 =3−(√8)=3−2(√2) ∣z_1 ∣−1=3+2(√2)−1=2+2(√2)>0 ∣z_2 ∣−1=3−2(√2)−1=2−2(√2) <0 ⇒ ∫_(∣z∣) f(z)dz=2iπ Res(f,z_2 ) f(z)=((2i)/((z−z_1 )(z−z_2 ))) ⇒ Res(f,z_2 )=((2i)/(z_2 −z_1 ))=((2i)/(−4(√2)))=−(i/(2(√2))) ∫_(∣z∣=1) f(z)dz=2iπ×((−i)/(2(√2))) =(π/( (√2))) ⇒★I=(π/( (√2)))★
$$ \\ $$$${I}=\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{1}+{sin}^{\mathrm{2}} {x}}=\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{1}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi} \frac{{dx}}{\mathrm{3}−{cos}\left(\mathrm{2}{x}\right)}\:\:\left(\mathrm{2}{x}={t}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{\mathrm{3}−{cost}}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{dt}}{\mathrm{3}−\frac{{e}^{{it}} +{e}^{−{it}} }{\mathrm{2}}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{\mathrm{2}{dt}}{\mathrm{6}−{e}^{{it}} −{e}^{−{it}\:\:} }\:\:\left({e}^{{it}} ={z}\right) \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\mathrm{6}−{z}−{z}^{−\mathrm{1}} \right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{i}\:{dz}}{\mathrm{6}{z}−{z}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{2}{idz}}{{z}^{\mathrm{2}} −\mathrm{6}{z}\:+\mathrm{1}} \\ $$$${f}\left({z}\right)=\frac{\mathrm{2}{i}}{{z}^{\mathrm{2}} −\mathrm{6}{z}+\mathrm{1}}\:\:{les}\:{poles}? \\ $$$$\Delta^{'} =\mathrm{3}^{\mathrm{2}} −\mathrm{1}=\mathrm{8}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\mathrm{3}+\sqrt{\mathrm{8}}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\mathrm{3}−\sqrt{\mathrm{8}}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}=\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}>\mathrm{0} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}=\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$<\mathrm{0}\:\Rightarrow \\ $$$$\int_{\mid{z}\mid} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left({f},{z}_{\mathrm{2}} \right) \\ $$$${f}\left({z}\right)=\frac{\mathrm{2}{i}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${Res}\left({f},{z}_{\mathrm{2}} \right)=\frac{\mathrm{2}{i}}{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }=\frac{\mathrm{2}{i}}{−\mathrm{4}\sqrt{\mathrm{2}}}=−\frac{{i}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {f}\left({z}\right){dz}=\mathrm{2}{i}\pi×\frac{−{i}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\Rightarrow\bigstar{I}=\frac{\pi}{\:\sqrt{\mathrm{2}}}\bigstar \\ $$
Answered by mnjuly1970 last updated on 18/Nov/23
I=2 ∫_0 ^( (π/2)) ((1+tan^( 2) (x))/(1+2tan^2 (x)))dx=^(tanx=y) = 2∫_0 ^( ∞) (dy/(1+2y^2 )) = ∫_0 ^( ∞) (dy/((((√2)/2))^2 +y^2 )) =[ (√2) tan^( −1) ( y(√2) )]_0 ^∞ =((π(√2))/2)
$$\:\:\mathrm{I}=\mathrm{2}\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{tan}^{\:\mathrm{2}} \left({x}\right)}{\mathrm{1}+\mathrm{2}{tan}^{\mathrm{2}} \left({x}\right)}{dx}\overset{{tanx}={y}} {=} \\ $$$$\:\:\:\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{{dy}}{\mathrm{1}+\mathrm{2}{y}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{dy}}{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:=\left[\:\sqrt{\mathrm{2}}\:{tan}^{\:−\mathrm{1}} \left(\:{y}\sqrt{\mathrm{2}}\:\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$

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