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Question Number 204275 by Frix last updated on 11/Feb/24
Show that ∫_0 ^(π/4) (√(tan x)) (√(1−tan x)) dx=(((√((√2)−1))/( (√2)))−1)π
$$\mathrm{Show}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\sqrt{\mathrm{tan}\:{x}}\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}\:{dx}=\left(\frac{\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}\right)\pi \\ $$
Answered by witcher3 last updated on 11/Feb/24
nice problem tan(x)=y ∫_0 ^1 ((√(y(1−y)))/(1+y^2 ))dy=I;(1/y)=z⇒I=∫_1 ^∞ ((√(z−1))/(z(z^2 +1)))dz (√(z−1))=x⇒I=∫_0 ^∞ ((2x^2 )/(((x^2 +1)^2 +1)(x^2 +1)))dx =∫_0 ^∞ ((2x^2 )/((x^2 +1+i)(x^2 +1−i)))=∫_(−∞) ^∞ ((x^2 dx)/((x^2 +1−i)(x^2 +1+i)(x^2 +1)))=I Residue Theorem overH {z∈C∣im(z)≥0} x^2 +1−i=0⇒x^2 =(√2)e^(3((iπ)/4)) x1=(√(√2))e^((i3π)/8) ∈H x^2 =−1−i⇒x^2 =(√2)e^((5iπ)/4) ⇒x_2 =(√(√2))e^((5iπ)/8) x^2 +1=0⇒x_3 =i I=2iπ {Res(f,z_1 )+Res(f,z_2 ))Resf(x_3 )} =2iπ{(z_1 ^2 /((2z_(1 ) )(z_1 ^2 +1+i)(z_1 ^2 +1)))+(z_2 ^2 /((2z_2 )(z_2 ^2 +1−i)(z_2 ^2 +1)))+(z_3 ^2 /(2z_3 (z_3 ^2 +1+i)(z_3 ^2 +1−i)))) =2iπ{(((√(√2))e^((3iπ)/8) )/(4i^2 ))+(((√(√2))e^((5iπ)/8) )/(−4))+(i/(2(i)(−i)))} =(i/2)π(√(√2))(−e^((3iπ)/8) −e^((5iπ)/8) )−π =(π/2)(√(√2))(2sin(3(π/8)))−π =π(√((√2).sin^2 (((3π)/8))))−π =π(√((1/2)(1−cos(3(π/4)))(√2)))−π =π(√((1/( (√2)))+(1/2)))−π =π(((√((√2)+1))/( (√2)))−1)
$$\mathrm{nice}\:\mathrm{problem} \\ $$$$\mathrm{tan}\left(\mathrm{x}\right)=\mathrm{y} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{y}\left(\mathrm{1}−\mathrm{y}\right)}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\mathrm{dy}=\mathrm{I};\frac{\mathrm{1}}{\mathrm{y}}=\mathrm{z}\Rightarrow\mathrm{I}=\int_{\mathrm{1}} ^{\infty} \frac{\sqrt{\mathrm{z}−\mathrm{1}}}{\mathrm{z}\left(\mathrm{z}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dz} \\ $$$$\sqrt{\mathrm{z}−\mathrm{1}}=\mathrm{x}\Rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2x}^{\mathrm{2}} }{\left(\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{i}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{i}\right)}=\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{i}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{i}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)}=\mathrm{I} \\ $$$$\mathrm{Residue}\:\mathrm{Theorem}\:\mathrm{overH}\:\left\{\mathrm{z}\in\mathbb{C}\mid\mathrm{im}\left(\mathrm{z}\right)\geqslant\mathrm{0}\right\} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{i}=\mathrm{0}\Rightarrow\mathrm{x}^{\mathrm{2}} =\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{3}\frac{\mathrm{i}\pi}{\mathrm{4}}} \\ $$$$\mathrm{x1}=\sqrt{\sqrt{\mathrm{2}}}\mathrm{e}^{\frac{\mathrm{i3}\pi}{\mathrm{8}}} \in\mathrm{H} \\ $$$$\mathrm{x}^{\mathrm{2}} =−\mathrm{1}−\mathrm{i}\Rightarrow\mathrm{x}^{\mathrm{2}} =\sqrt{\mathrm{2}}\mathrm{e}^{\frac{\mathrm{5i}\pi}{\mathrm{4}}} \Rightarrow\mathrm{x}_{\mathrm{2}} =\sqrt{\sqrt{\mathrm{2}}}\mathrm{e}^{\frac{\mathrm{5i}\pi}{\mathrm{8}}} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{x}_{\mathrm{3}} =\mathrm{i} \\ $$$$\left.\mathrm{I}=\mathrm{2i}\pi\:\left\{\mathrm{Res}\left(\mathrm{f},\mathrm{z}_{\mathrm{1}} \right)+\mathrm{Res}\left(\mathrm{f},\mathrm{z}_{\mathrm{2}} \right)\right)\mathrm{Resf}\left(\mathrm{x}_{\mathrm{3}} \right)\right\} \\ $$$$=\mathrm{2i}\pi\left\{\frac{\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} }{\left(\mathrm{2z}_{\mathrm{1}\:} \right)\left(\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}+\mathrm{i}\right)\left(\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{z}_{\mathrm{2}} ^{\mathrm{2}} }{\left(\mathrm{2z}_{\mathrm{2}} \right)\left(\mathrm{z}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{1}−\mathrm{i}\right)\left(\mathrm{z}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{z}_{\mathrm{3}} ^{\mathrm{2}} }{\mathrm{2z}_{\mathrm{3}} \left(\mathrm{z}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{1}+\mathrm{i}\right)\left(\mathrm{z}_{\mathrm{3}} ^{\mathrm{2}} +\mathrm{1}−\mathrm{i}\right)}\right) \\ $$$$=\mathrm{2i}\pi\left\{\frac{\sqrt{\sqrt{\mathrm{2}}}\mathrm{e}^{\frac{\mathrm{3i}\pi}{\mathrm{8}}} }{\mathrm{4i}^{\mathrm{2}} }+\frac{\sqrt{\sqrt{\mathrm{2}}}\mathrm{e}^{\frac{\mathrm{5i}\pi}{\mathrm{8}}} }{−\mathrm{4}}+\frac{\mathrm{i}}{\mathrm{2}\left(\mathrm{i}\right)\left(−\mathrm{i}\right)}\right\} \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\pi\sqrt{\sqrt{\mathrm{2}}}\left(−\mathrm{e}^{\frac{\mathrm{3i}\pi}{\mathrm{8}}} −\mathrm{e}^{\frac{\mathrm{5i}\pi}{\mathrm{8}}} \right)−\pi \\ $$$$=\frac{\pi}{\mathrm{2}}\sqrt{\sqrt{\mathrm{2}}}\left(\mathrm{2sin}\left(\mathrm{3}\frac{\pi}{\mathrm{8}}\right)\right)−\pi \\ $$$$=\pi\sqrt{\sqrt{\mathrm{2}}.\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}−\pi \\ $$$$=\pi\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\left(\mathrm{3}\frac{\pi}{\mathrm{4}}\right)\right)\sqrt{\mathrm{2}}}−\pi \\ $$$$=\pi\sqrt{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}}}−\pi \\ $$$$=\pi\left(\frac{\sqrt{\sqrt{\mathrm{2}}+\mathrm{1}}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Frix last updated on 11/Feb/24
Thank you!
$$\mathrm{Thank}\:\mathrm{you}! \\ $$
Commented by witcher3 last updated on 11/Feb/24
withe pleasur
$$\mathrm{withe}\:\mathrm{pleasur} \\ $$

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