Question

Solve:
$I=\int \frac{dx}{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\sqrt{3}}$

Answer 1

$\begin{array}{c}Wehave\\ I=\int \frac{dx}{\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\sqrt{3}}\\ =\int \frac{2\tan \frac{x}{2}}{1-{\tan }^2\frac{x}{2}}\\ =\int \frac{{\sec }^2\frac{x}{2}dx}{1-{\tan }^2\frac{x}{2}+2\sqrt{3}\tan \frac{x}{2}}\\ put\tan \frac{x}{2}=t\\ \therefore dt=\frac{1}{2}se{x}^2\frac{x}{2}dx\\ Now\\ I=\int \frac{2dt}{1+2\sqrt{2}t-t^2}\\ =\int \frac{2dt}{\left(t-\sqrt{3}-2\right)\left(t-\sqrt{3}+2\right)}\\ =\frac{1}{2}\int \left(\frac{1}{t-\sqrt{3}-2}-\frac{1}{t-\sqrt{3}+2}\right)dt\\ =\frac{1}{2}\int \frac{dt}{\left(t-\sqrt{3}-2\right)}-\frac{1}{2}\int \frac{dt}{t-\sqrt{3}+2}\\ =\frac{1}{2}\ln \left[t-\sqrt{3}0-2\right]-\frac{1}{2}\ln \left[t-\sqrt{3}-2\right]\end{array}$