A Definite Integral I

Question

Given the definite integral \begin{align} \int_{0}^{\pi} \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt = -2\pi \end{align} then what is the general value of the indefinite integral \begin{align} \int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt \,\,\, ? \end{align}

Answer 1

$$\begin{align} \int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt &= \int\sec^2t\ln\left(1+\sin^2t\right) dt\\ &=\ln(1+\sin^2t)\int\sec^2tdt-\int\frac{2\sin t\cos t}{1+\sin^2t}\int\sec^2tdtdt\\ &=\ln(1+\sin^2t)\tan t-\int\frac{2\sin t\cos t\tan t}{1+\sin^2t}dt\\ &=\ln(1+\sin^2t)\tan t-2\int\left(1-\frac{1}{1+\sin^2t}\right)dt\\ &=\ln(1+\sin^2t)\tan t-2t+2\int\frac1{1+\sin^2t}dt \end{align}$$ Now $$\begin{align} \int\frac1{1+\sin^2t}dt &=\int\frac{\sec^2tdt}{\sec^2t+\tan^2 t}\\ &\stackrel{u=\tan x}\equiv\int\frac{du}{2+u^2}\\ &=\frac1{\sqrt2}\arctan\frac{\tan x}{\sqrt2}+c \end{align}$$ So $$ \int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt =\ln(1+\sin^2t)\tan t-2t+\sqrt2\arctan\left(\frac{\tan x}{\sqrt2}\right)+C$$ Also $$\left( \ln(1+\sin^2t)\tan t-2t+\sqrt2\arctan\left(\frac{\tan x}{\sqrt2}\right)\right)_0^{\pi} \\=-2(\pi-0)+\sqrt2\left(\arctan(0)-\arctan(0)\right)+\ln(1+0)\tan(\pi)-\ln(1+0)\tan(0) \\=-2\pi$$

Answer 2

$$I=\int \frac{1+\cot^{2}(t)}{\cot^{2}(t)} \, \ln\left( \frac{1+2\tan^{2}(t)}{1+\tan^{2}(t)} \right) \, dt \,\,\, $$

$$=\int(1+\tan^2t)[\ln(1+2\tan^2t)-\ln(1+\tan^2t)]dt$$

Setting $\tan t=u,$

$$I=\int[\ln(1+2u^2)-\ln(1+u^2)]du$$

which should be addressed by Integration by Parts