Prove-that-0-pi-2-tan-2x-sin-4-x-4cos-2-x-cos-4-x-4sin-2-x-1-

Question Number 178693 by peter frank last updated on 20/Oct/22

Prove that ∫_0 ^(π/2) ((tan 2x)/( (√(sin^4 x+4cos^2 x)) −(√(cos^4 x+4sin^2 x))))=1

$$\mathrm{P}{rove}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{tan}\:\mathrm{2x}}{\:\sqrt{\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+\mathrm{4cos}\:^{\mathrm{2}} \mathrm{x}}\:−\sqrt{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}+\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}}}=\mathrm{1} \\ $$

Commented by peter frank last updated on 20/Oct/22

$$\mathrm{1} \\ $$

Commented by som(math1967) last updated on 20/Oct/22

$$\mathrm{1}\:{or}\:−\mathrm{1}\:? \\ $$

Commented by MJS_new last updated on 20/Oct/22

I don′t think the integral exists. ∫_0 ^(π/2) ((tan 2x)/(cos 2x))dx=2∫_0 ^(π/4) ((tan 2x)/(cos 2x))dx which does not converge

$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{exists}. \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{\mathrm{tan}\:\mathrm{2}{x}}{\mathrm{cos}\:\mathrm{2}{x}}{dx}=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\frac{\mathrm{tan}\:\mathrm{2}{x}}{\mathrm{cos}\:\mathrm{2}{x}}{dx}\:\mathrm{which}\:\mathrm{does}\:\mathrm{not}\:\mathrm{converge} \\ $$

Commented by Frix last updated on 21/Oct/22

you are right, I did not think of this, changed my answer

$$\mathrm{you}\:\mathrm{are}\:\mathrm{right},\:\mathrm{I}\:\mathrm{did}\:\mathrm{not}\:\mathrm{think}\:\mathrm{of}\:\mathrm{this},\:\mathrm{changed} \\ $$$$\mathrm{my}\:\mathrm{answer} \\ $$

Commented by Ar Brandon last updated on 21/Oct/22

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Commented by Rasheed.Sindhi last updated on 21/Oct/22

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Answered by Frix last updated on 21/Oct/22

sin^4 x +4cos^2 x =(1+cos^2 x)^2 cos^4 x +4sin^2 x =(1+sin^2 x)^2 1+cos^2 x −(1+sin^2 x)=cos 2x ∫((tan 2x)/( (√(sin^4 x +4cos^2 x))−(√(cos^4 x +4sin^2 x))))dx= =∫((tan 2x)/(cos 2x))dx=(1/(2cos 2x))+C but the integral doesn′t exist within 0≤x≤(π/2)

$$\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{4cos}^{\mathrm{2}} \:{x}\:=\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{cos}^{\mathrm{4}} \:{x}\:+\mathrm{4sin}^{\mathrm{2}} \:{x}\:=\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}\:−\left(\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}\right)=\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\int\frac{\mathrm{tan}\:\mathrm{2}{x}}{\:\sqrt{\mathrm{sin}^{\mathrm{4}} \:{x}\:+\mathrm{4cos}^{\mathrm{2}} \:{x}}−\sqrt{\mathrm{cos}^{\mathrm{4}} \:{x}\:+\mathrm{4sin}^{\mathrm{2}} \:{x}}}{dx}= \\ $$$$=\int\frac{\mathrm{tan}\:\mathrm{2}{x}}{\mathrm{cos}\:\mathrm{2}{x}}{dx}=\frac{\mathrm{1}}{\mathrm{2cos}\:\mathrm{2}{x}}+{C} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist}\:\mathrm{within}\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$

Commented by Frix last updated on 20/Oct/22

$$\mathrm{yes},\:\mathrm{I}\:\mathrm{made}\:\mathrm{a}\:\mathrm{typo} \\ $$

Commented by som(math1967) last updated on 20/Oct/22

sir ((tan2x)/(cos2x))=tan2xsec2x ∫_0 ^(π/2) tan2xsec2xdx =(1/2)[sec2x]_0 ^(π/2) =(1/2)×(−2)=−1 am i correct?

$${sir}\:\frac{{tan}\mathrm{2}{x}}{{cos}\mathrm{2}{x}}={tan}\mathrm{2}{xsec}\mathrm{2}{x}\: \\ $$$$\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}\mathrm{2}{xsec}\mathrm{2}{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{sec}\mathrm{2}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}}×\left(−\mathrm{2}\right)=−\mathrm{1} \\ $$$$\:{am}\:{i}\:{correct}? \\ $$

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