Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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)))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com