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How to evaluate this given expression? $$\int\frac{du}{\sqrt{9e^{-2u}-1}}$$ I got so many tries but I'm not sure of my answer because somebody said that it was wrong, they told me that I used a wrong formula applied! That's why I ask a support here I want correct explanation and answer of this given!

Thanks!

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    $\begingroup$ Write down your answer, let us differentiate it and we shall check who's wrong: you or that somebody. $\endgroup$ – DonAntonio Jan 31 '18 at 9:41
  • $\begingroup$ My first answer was arcsin(e^u/3)+C. But my professor told me it should be -arcsec(e^-u/3) + C. Other said that it was -arctangent($$\sqrt{(9e^-2u)-1})+C. Which one is correct? $\endgroup$ – Duane Nestor Jr. Jan 31 '18 at 9:54
  • $\begingroup$ $$\left(\arcsin\frac{e^u}3\right)'=\frac{e^u}3\cdot\frac1{\sqrt{1-\frac{e^{2u}}9}}=\frac{e^u}{\sqrt{9-e^{2u}}}=\frac1{\sqrt{9e^{-2u}-1}}$$and thus your answer is correct... $\endgroup$ – DonAntonio Jan 31 '18 at 10:26
  • $\begingroup$ In addition to @DonAntonio's comment, the arctan answer is also correct. But the arcsec one is not. $\endgroup$ – mickep Jan 31 '18 at 10:27
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try to substitute $\frac{1}{3}e^{u} = cost$ then $du = -tg(t)dt$. And the integral will be $\int \frac{-tgtdt}{\sqrt{\frac{1}{cos(t)^2} - 1}}$ then use that $(cos(t))^2 + (sin(t))^2 = 1$

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Considering $$I=\int\frac{du}{\sqrt{9e^{-2u}-1}}$$ use $$\sqrt{9e^{-2u}-1}=t\implies u=-\frac{1}{2} \log \left(\frac{1}{9} \left(t^2+1\right)\right)\implies du=-\frac{t}{t^2+1}\,dt$$ This makes $$I=-\int\frac{dt}{t^2+1}$$

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