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In the answer, author uses the principal square root of $1 + \tan^2 x$ to evaluate the integral $\displaystyle \int {\sec^2 x \over (\sec x + \tan x)^{9/2}}$. Author rewrites $\sec x $ as $\sqrt{1 + \tan^2 x}$ and proceeds with it.

Since secant of an angle can be negative in second and third quadrant, I want to know why does using principal square root is justified here ?

  • I don't understand what you mean to ask? Clearly $1+tan^2x$ is $sec^2x$ and root of it is $+-secx$ isn't it? – Iti Shree Apr 22 '17 at 20:10
  • The author used principal square root. See https://math.stackexchange.com/questions/2242711/is-it-wrong-to-say-that-100-is-solution-of-sqrt-x10-0 for more information. –  Apr 22 '17 at 20:12
  • ^Okay I got what you're asking for I'll check it out now. – Iti Shree Apr 22 '17 at 20:16
  • I think it's only justified if $x$ is restricted to an interval where $\cos x$ is positive. – user232456 Apr 23 '17 at 18:40
  • @user232456 Check the original question, I don't think indefinite integrals are restricted to some interval. –  Apr 23 '17 at 18:48

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