Menu Close

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
1
$$\mathrm{1} \\ $$
Commented by som(math1967) last updated on 20/Oct/22
1 or −1 ?
$$\mathrm{1}\:{or}\:−\mathrm{1}\:? \\ $$
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
��
Commented by Rasheed.Sindhi last updated on 21/Oct/22
��
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
yes, I made a typo
$$\mathrm{yes},\:\mathrm{I}\:\mathrm{made}\:\mathrm{a}\:\mathrm{typo} \\ $$
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?
$${sir}\:\frac{{tan}\mathrm{2}{x}}{{cos}\mathrm{2}{x}}={tan}\mathrm{2}{xsec}\mathrm{2}{x}\: \\ $$$$\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {tan}\mathrm{2}{xsec}\mathrm{2}{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[{sec}\mathrm{2}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}}×\left(−\mathrm{2}\right)=−\mathrm{1} \\ $$$$\:{am}\:{i}\:{correct}? \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *