integral and sines and stuff 2

Note that $\tanh^{-1}(\tan x) \; = \; \tfrac12\ln\left(\frac{1 + \tan x}{1 - \tan x}\right) \; = \; \tfrac12\ln\left(\frac{\cos x+ \sin x}{\cos x - \sin x}\right) \; = \; \tfrac12\ln\big(\cot(\tfrac14\pi - x)\big)$ and so $\begin{aligned} \int_0^{\frac14\pi}\tanh^{-1}(\tan x)\,dx & = \; \tfrac12\int_0^{\frac14\pi}\ln\big(\cot(\tfrac14\pi - x)\big)\,dx \; = \; \tfrac12\int_0^{\frac14\pi}\ln(\cot x)\,dx \\ & = \; -\tfrac12\int_0^{\frac14\pi}\ln(\tan x)\,dx \; = \; -\tfrac12\int_0^1 \frac{\ln u}{1 + u^2}\,du \; = \; \tfrac12\sum_{n=0}^\infty (-1)^{n-1}\int_0^1 u^{2n} \ln u\,du \\ & = \; \tfrac12\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \; = \; \tfrac12G \end{aligned}$ making the answer $\boxed{2}$ .

A denominator cannot be $0$ , so we can exclude 0 from the answer.

Next, the answer of integral cannot be negative.

So if I let the answer to $A$ , then $a > 0$ .

And the answer of integral can be positive and irrational, rational.

We get $\displaystyle \int ^ { \frac { \pi } { 4 } } _ { 0 } \tanh ^ { -1 } ( \tan ( x ) ) dx = \frac { G } { a }$ .

Solving this problem, we get $a = \boxed { \boxed { 2 } }$ .