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Evaluate $$\int_{0}^{\frac{\pi}{4}}\frac{\sec^2 \theta }{(1-\tan \theta )}\ d \theta$$

Here's my attempt:

$$u=1-tan \theta \implies -du=\sec^2 \theta d \theta$$

Substituting back in, I get this: $$-\int_{0}^{\frac{\pi}{4}} \frac{du}{u}$$

Integrating I get this (I prefer to not change my bounds and to back-substitute at the end):

$$-ln\lvert u \rvert$$ evaluated from $0$ to $\frac{\pi}{4}$

back-substituting gives me this:

$$-ln \lvert 1-tan \theta \rvert$$ evaluated from $0$ to $\frac{\pi}{4}$

Plugging in the bounds, I get this:

$$-\left[\left(ln\lvert1-tan \left(\frac{\pi}{4}\right)\rvert \right)-\left(ln\lvert1-tan\left(0\right)\rvert \right)\right]$$

Which gives me this:

$$=-[0-0]=0$$

Now I know that I must have messed up somewhere because, looking at the graph of $\sec^2$, I see that it approaches $\infty$ as $x$ approaches $\frac{\pi}{4}$, so the area must be infinite and so I should be getting a divergent result. Can someone show me where and how I messed up?

Thanks in advance

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    $\begingroup$ Note, when you change the variable you should change the bounds... $\endgroup$ – Thomas Andrews Jun 28 '15 at 14:30
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    $\begingroup$ One clear way to deal with what Thomas Andrews said without having to update for $u$ bounds is to say "evaluated from $\theta=0$ to $\theta=\pi/4$". This is a pretty good habit for multivariable calculus as well. $\endgroup$ – Ian Jun 28 '15 at 14:32
  • $\begingroup$ By the way, +1 for showing your work. It made it very easy to help you. Next time try to use MathJax. $\endgroup$ – Ian Jun 28 '15 at 15:17
  • $\begingroup$ No problem... thanks you for being so responsive and quick! And I thought mathjax was what I was using...? $\endgroup$ – Keenan Jun 28 '15 at 15:52
  • $\begingroup$ Is there an online equation editor or something similar to this (codecogs.com/latex/eqneditor.php) for MathJax? $\endgroup$ – Keenan Jun 28 '15 at 16:01
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Your problem is right at the end. You have

$$-\ln(|1-\tan(\pi/4)|)+\ln(|1-0|)"="-\ln(0)+\ln(1)$$

where I put the equals sign in scare quotes because that's what you get when you try to substitute, but $\ln(0)$ is not defined. When you rephrase in terms of an improper integral (as you should, since your original integrand blows up at $\pi/4$), you get divergence as you anticipated.

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  • $\begingroup$ Ahh I see... When you say "rephrase in terms of an improper integral" do you mean like rewrite it in a limit sense, with one of the bounds replaced with "b"? $\endgroup$ – Keenan Jun 28 '15 at 14:36
  • $\begingroup$ @Keenan Yes. Your original integral should be written as $\lim_{b \to \pi/4^-} \int_0^b \frac{\sec^2(\theta)}{1-\tan(\theta)} d \theta$. When you substitute you then get $\lim_{a \to 0^+} -\int_1^a \frac{1}{u} du$. $\endgroup$ – Ian Jun 28 '15 at 14:38

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