Question-132554

Question Number 132554 by mohammad17 last updated on 15/Feb/21

Commented by MJS_new last updated on 15/Feb/21

we had this many times before. simply substitute t=(√(tan θ)) and there you go

$$\mathrm{we}\:\mathrm{had}\:\mathrm{this}\:\mathrm{many}\:\mathrm{times}\:\mathrm{before}.\:\mathrm{simply} \\ $$$$\mathrm{substitute}\:{t}=\sqrt{\mathrm{tan}\:\theta}\:\mathrm{and}\:\mathrm{there}\:\mathrm{you}\:\mathrm{go} \\ $$

Commented by liberty last updated on 15/Feb/21

I=∫_0 ^(π/2) (√(tan x)) dx = ∫_0 ^(π/2) (√(cot x)) dx 2I=∫_0 ^(π/2) ((sin x+cos x)/( (√(sin xcos x)))) dx let u = sin x−cos x ⇒du=(sin x+cos x) dx { ((u=1)),((u=−1)) :} ⇒u^2 =1−2sin xcos x sin xcos x = ((1−u^2 )/2) 2I=∫_(−1) ^1 (((√2) du)/( (√(1−u^2 )))) I=(1/( (√2)))∫ (du/( (√(1−u^2 )))) I=(1/( (√2)))[ arcsin (u)]_(−1) ^1 = (1/( (√2))) [ (π/2)+(π/2)] = (π/( (√2)))

$$\mathrm{I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \sqrt{\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\:=\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \sqrt{\mathrm{cot}\:\mathrm{x}}\:\mathrm{dx} \\ $$$$\mathrm{2I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\:\sqrt{\mathrm{sin}\:\mathrm{xcos}\:\mathrm{x}}}\:\mathrm{dx} \\ $$$$\mathrm{let}\:\mathrm{u}\:=\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\:\Rightarrow\mathrm{du}=\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)\:\mathrm{dx} \\ $$$$\:\begin{cases}{\mathrm{u}=\mathrm{1}}\\{\mathrm{u}=−\mathrm{1}}\end{cases}\:\Rightarrow\mathrm{u}^{\mathrm{2}} =\mathrm{1}−\mathrm{2sin}\:\mathrm{xcos}\:\mathrm{x} \\ $$$$\:\mathrm{sin}\:\mathrm{xcos}\:\mathrm{x}\:=\:\frac{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{2I}=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{2}}\:\mathrm{du}}{\:\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\:\frac{\mathrm{du}}{\:\sqrt{\mathrm{1}−\mathrm{u}^{\mathrm{2}} }}\: \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\:\mathrm{arcsin}\:\left(\mathrm{u}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} =\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left[\:\frac{\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\right] \\ $$$$=\:\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$

Answered by Ñï= last updated on 15/Feb/21

∫_0 ^(π/2) (√(tanx))dx =∫_0 ^(π/2) sin^(1/2) xcos^(−(1/2)) xdx =(1/2)B(((1+(1/2))/2),((1−(1/2))/2)) =(1/2)Γ((3/4))Γ((1/4)) =(π/(2sin(π/4))) =(π/( (√2)))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{{tanx}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\frac{\mathrm{1}}{\mathrm{2}}} {xcos}^{−\frac{\mathrm{1}}{\mathrm{2}}} {xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{B}\left(\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}},\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}{sin}\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$

Answered by Dwaipayan Shikari last updated on 15/Feb/21

∫_0 ^(π/2) sin^(1/2) x cos^(−(1/2)) xdx=((Γ((3/4))Γ((1/4)))/(2Γ(1)))=(π/( (√2)))

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\frac{\mathrm{1}}{\mathrm{2}}} {x}\:{cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} {xdx}=\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\mathrm{2}\Gamma\left(\mathrm{1}\right)}=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$

Answered by mathmax by abdo last updated on 15/Feb/21

I=∫_0 ^(π/2) (√(tanθ))dθ we do the changement (√(tanθ))=t ⇒tanθ=t^2 ⇒ θ=arctan(t^2 ) ⇒I =∫_0 ^∞ t.((2t)/(1+t^4 ))dt =2∫_0 ^∞ (t^2 /(1+t^4 ))dt =_(t^4 =z) 2.(1/4)∫_0 ^∞ (((z^(1/4) )^2 )/(1+z))z^((1/4)−1) dz =(1/2)∫_0 ^∞ (z^((1/2)+(1/4)−1) /(1+z))dz =(1/2)∫_0 ^∞ (z^((3/4)−1) /(1+z)) =(1/2) ×(π/(sin(((3π)/4)))) =(π/(2.((√2)/2)))=(π/( (√2)))

$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \sqrt{\mathrm{tan}\theta}\mathrm{d}\theta\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\sqrt{\mathrm{tan}\theta}=\mathrm{t}\:\Rightarrow\mathrm{tan}\theta=\mathrm{t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\theta=\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{t}.\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=_{\mathrm{t}^{\mathrm{4}} \:=\mathrm{z}} \:\:\:\mathrm{2}.\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)^{\mathrm{2}} }{\mathrm{1}+\mathrm{z}}\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+\mathrm{z}}\mathrm{dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+\mathrm{z}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:×\frac{\pi}{\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{2}.\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$

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