Easy Calculus

$\displaystyle \int\tan(\dfrac{\theta}{2}) d\theta = -2\text{In(}\cos(\dfrac{\theta}{2})) \\ \displaystyle \int_0^{\dfrac{\pi}{2}} \tan(\dfrac{\theta}{2}) \mathrm{d}\theta\\ \displaystyle = -2\text{ln}(\cos(\dfrac{\pi}{4})) - (- 2\text{In}(\cos(0)))\\ \displaystyle = -2\text{In}(\dfrac{1}{\sqrt[]{2}}) + 2\text{In}(1)\\ \displaystyle = -2\text{In}((2)^{\dfrac{-1}{2}}) + 0\\ \displaystyle = -2 \times \dfrac{-1}{2} \text{In}(2) \\ \displaystyle = \text{In(2)} \\ \displaystyle \boxed{\therefore A = 2}\\ \text{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \displaystyle I = \int\tan(\dfrac{\theta}{2}) d\theta = ? \\ \displaystyle I = \int\dfrac{\sin\left(\dfrac{\theta}{2}\right)}{\cos\left(\dfrac{\theta}{2}\right)} d\theta \\ \displaystyle \text{Let, } x = \dfrac{\theta}{2} \\ \displaystyle \dfrac{dx}{d\theta} = \dfrac{d}{d\theta} \dfrac{\theta}{2} \\ \displaystyle d\theta = 2dx \\ \displaystyle I = \int\dfrac{\sin(x)}{\cos(x)} ~ 2 ~ dx \\ \displaystyle = 2 \int\dfrac{\sin(x)}{\cos(x)} ~ dx \\ \displaystyle \text{Let, } a = \cos(x) \\ \displaystyle dx = -\dfrac{da}{\sin(x)} \\ \displaystyle I = 2 \int\dfrac{\sin(x)}{a} ~\times -\dfrac{da}{\sin(x)} \\ \displaystyle = -2 \int \dfrac{1}{a} ~ da \\ \displaystyle = -2 \text{In}(|a|)\\ \displaystyle = -2 \text{In}(|\cos(x)|) \\ \displaystyle \therefore \int\tan(\dfrac{\theta}{2}) d\theta = -2 \text{In}(|\cos(\dfrac{\theta}{2})|)$

$\begin{aligned} I & = \int_0^\frac \pi 2 \tan \frac \pi 2 \ d\theta \\ & = \int_0^\frac \pi 2 \frac {\sin \frac \pi 2}{\cos \frac \pi 2} \ d\theta & \small \color{#3D99F6}{\text{Let }x = \cos \frac \pi 2, \ dx = - \frac 12 \sin \frac \pi 2 \ d\theta} \\ & = - 2 \int^{\frac 1{\sqrt 2}}_1 \frac {dx}x \\ & = -2 \ln x \bigg|^{\frac 1{\sqrt 2}}_1 \\ & = -2 \left(\ln \left(\frac 1{\sqrt 2} \right) - \ln 1\right) \\ & = \ln 2 \end{aligned}$

$\implies A = \boxed{2}$