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Question Number 94718 by john santu last updated on 20/May/20

∫ (√(tan x)) dx = ∫(((√(tan x))+(√(cot x)))/2) dx + ∫ (((√(tan x))−(√(cot x)))/2) dx =(1/((√2) ))∫ ((sin x+cos x)/(√(sin 2x))) dx + (1/((√2) ))∫ ((sin x−cos x)/(√(sin 2x))) dx = (1/((√2) ))∫ ((sin x+cos x)/(√(1−(sin x−cos x)^2 ))) dx + (1/(√2)) ∫ ((sin x−cos x)/(√((sin x+cos x)^2 −1))) dx = (1/((√2) ))∫ (dt/(√(1−t^2 ))) +(1/((√2) ))∫ ((−du)/(√(u^2 −1))) = (1/((√2) ))sin^(−1) (t) −(1/(√2)) ln(u+(√(u^2 −1))) +c = (1/(√2)) sin^(−1) (sin x−cos x)− (1/(√2)) ln (sin x+cos x+(√(sin 2x))) + c where t = sin x−cos x ; u = sin x+cos x

$$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\:= \\ $$$$\int\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}+\sqrt{\mathrm{cot}\:\mathrm{x}}}{\mathrm{2}}\:\mathrm{dx}\:+\:\int\:\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}−\sqrt{\mathrm{cot}\:\mathrm{x}}}{\mathrm{2}}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\sqrt{\mathrm{sin}\:\mathrm{2x}}}\:\mathrm{dx}\:+\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\sqrt{\mathrm{sin}\:\mathrm{2x}}}\:\mathrm{dx} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\sqrt{\mathrm{1}−\left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} }}\:\mathrm{dx}\:+\: \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\int\:\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\sqrt{\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\right)^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx}\: \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{\mathrm{dt}}{\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\:+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\int\:\frac{−\mathrm{du}}{\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}\:}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{t}\right)\:−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{ln}\left(\mathrm{u}+\sqrt{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}\right)\:+\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\right)− \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{ln}\:\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}+\sqrt{\mathrm{sin}\:\mathrm{2x}}\right)\:+\:\mathrm{c}\: \\ $$$$\mathrm{where}\:\mathrm{t}\:=\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}\:;\: \\ $$$$\mathrm{u}\:=\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}\: \\ $$

Commented by i jagooll last updated on 20/May/20

waww simple

$$\mathrm{waww}\:\mathrm{simple} \\ $$

Commented by MJS last updated on 20/May/20

nice but I think the easiest way to solve is t=(√(tan x)) or t=(√(cot x)); they lead to 2∫(t^2 /(t^4 +1))dt or −2∫(dt/(t^4 +1))

$$\mathrm{nice}\:\mathrm{but}\:\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{easiest}\:\mathrm{way}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{is} \\ $$$${t}=\sqrt{\mathrm{tan}\:{x}}\:\mathrm{or}\:{t}=\sqrt{\mathrm{cot}\:{x}};\:\mathrm{they}\:\mathrm{lead}\:\mathrm{to} \\ $$$$\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}{dt}\:\mathrm{or}\:−\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{1}} \\ $$$$ \\ $$

Commented by peter frank last updated on 20/May/20

good

$$\mathrm{good} \\ $$

Commented by peter frank last updated on 20/May/20

good

$$\mathrm{good} \\ $$

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