Evaluate $\, \displaystyle \int _{-1}^{-1/2} \frac{dx}{\sqrt {4x^2-1}}$.
$My\ work:-$
substituting $\, \displaystyle 2x=\sec (\theta),\, \,$
$\Rightarrow \displaystyle \displaystyle \int \frac{dx}{\sqrt{(2x)^2-1}} \displaystyle = \displaystyle \frac{1}{2}\int \frac{ \sec (\theta )\, \tan (\theta )\, d\theta }{\sqrt{\sec ^2(\theta )-1}} $
$\Rightarrow \displaystyle \int \frac{ \sec (\theta )\, \tan (\theta )\, }{2\, |\tan (\theta )|} \, d\theta$
now since$\displaystyle \, -1\leq 2x = \sec (\theta ) \leq -\frac{1}{2}\ $ we have quadrant II and III where $\sec (\theta )\leq 0.\,$
$Question:-$
1. In solution they given a diagram for IIIrd quadrant and Yes i know when we calculate sec we get same 2x, but i am confused how'd they come up with this diagram ?
and here's my Diagram, it will really helpful if someone please point out mistakes in my diagram
sorry for such Naive question
You have made mistake in your second step . You denoted $ 2x = \sec { \theta}$ . As $x \ge {-1} $, $2x \ge {-2}$ , therefore , $\sec { \theta} \ge {-2}$ . Therefore , $\cos { \theta} \le {{-1} \over 2}$. Now proceed.