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.
Best Answer
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?