Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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?

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com