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I'm working through an integral suggested for practice at the end of the Wikipedia article on differentiation under the integral sign, and I'm stuck.

I am attempting to evaluate this integral:

$$\int_0^{\pi/2} \frac{x}{\tan x} \ dx.$$

The article suggests the following parameterization:

$$F(a)=\int_0^{\pi/2} \frac{\tan^{-1}(a\tan (x))}{\tan x} \ dx.$$

Differentiating with respect to $a$, we get

$$F'(a)=\int_0^{\pi/2} \frac{1}{1+a^2\tan^2 x} \ dx.$$

I can't find a way to evaluate this, and neither can Wolfram Alpha. The special values $a=0,1$ are easy, but I fail to see how they help.

How can I finish evaluating this integral?

Edit: I think it's just substitution, maybe. I'll update the post accordingly soon.

Edit 2: Indeed, the substitution $u=\tan x$ and identity $\sec^2 = 1 + \tan^2$ transform the above integral into

$$F'(a) = \int_0^{\pi/2} \frac{1}{(1+u^2)(1+a^2u^2)}.$$

This can be solved with partial fractions.

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    $\begingroup$ What happened to posting the solutions to all of Rudin's exercises? $\endgroup$ – William Jul 5 '12 at 3:15
  • $\begingroup$ Is that a serious question? $\endgroup$ – Potato Jul 5 '12 at 3:15
  • $\begingroup$ I am actually thinking about possibly posting the solution to all the exercises from some textbook sometime in the future. I even asked the following meta question meta.math.stackexchange.com/questions/4337/… I wanted to see how that was going for you. $\endgroup$ – William Jul 5 '12 at 3:19
  • $\begingroup$ It was never my intent to write up all of them, just a few. I wanted to review a few concepts. $\endgroup$ – Potato Jul 5 '12 at 3:20
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    $\begingroup$ Possible duplicate (even though none of the answers use diff. under the integral): math.stackexchange.com/questions/166221/… $\endgroup$ – Joe Jul 5 '12 at 4:07
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You have that

$$\frac{1}{{\left( {1 + {u^2}} \right)\left( {1 + {a^2}{u^2}} \right)}} = \left( {\frac{A}{{1 + {u^2}}} + \frac{B}{{1 + {a^2}{u^2}}}} \right)$$

Thus you want (after cross mult.)

$$1 = A + A{a^2}{u^2} + B + B{u^2}$$

This is

$$\eqalign{ & A + B = 1 \cr & A{a^2} + B = 0 \cr} $$

Which gives

$$A = \frac{1}{{1 - {a^2}}}$$

and in turn

$$B = 1 - A = \frac{{{a^2}}}{{{a^2} - 1}}$$

which means

$$\frac{1}{{\left( {1 + {u^2}} \right)\left( {1 + {a^2}{u^2}} \right)}} = \frac{1}{{{a^2} - 1}}\left( {\frac{{{a^2}}}{{1 + {a^2}{u^2}}} - \frac{1}{{1 + {u^2}}}} \right)$$

Can you move on?

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  • $\begingroup$ Yes, I've done it out in full. I just missed the substitution I noted in my original post. $\endgroup$ – Potato Jul 5 '12 at 3:47
  • $\begingroup$ @Potato I'll delete this then, I guess. $\endgroup$ – Pedro Tamaroff Jul 5 '12 at 3:47
  • $\begingroup$ It's nice to have around. Please don't. $\endgroup$ – Potato Jul 5 '12 at 3:48
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Mathematica can do this integral just fine:

$$F'(a) = \frac{a \tan^{-1}(a \tan x)-\tan^{-1}(\tan x)}{a^2-1} \Bigg|_{0}^{\tfrac{\pi}{2}} = \frac{\pi}{2+2a}.$$

Hence the desired integral is: $$F(1) = F(0) +\int_{0}^{1} \! \frac{\pi}{2+2a}\,\mathrm{d}a =\frac{\pi}{2} \log(2).$$

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