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Establish relations for the integral from 0 to π/4 of tann x

by RoRi

Define the function

    \[ f(n) = \int_0^{\frac{\pi}{4}} \tan^n x \, dx. \]

Establish the following:

  1. \displaystyle{f(n+1) < f(n)}.
  2. \displaystyle{f(n)+f(n-2) = \frac{1}{n-1}} if n>2.
  3. \displaystyle{\frac{1}{n+1} < 2f(n) < \frac{1}{n-1}} if n > 2.
  1. Proof. We consider f(n) - f(n+1),

        \begin{align*} f(n) - f(n+1) &= \int_0^{\frac{\pi}{4}} \tan^n x \, dx - \int_0^{\frac{\pi}{4}} \tan^{n+1} x \, dx \\[9pt] &= \int_0^{\frac{\pi}{4}} (\tan^n x - \tan^{n+1} x) \, dx \\[9pt] &= \int_0^{\frac{\pi}{4}} \tan^n x (1 - \tan x) \, dx. \end{align*}

    This is then greater than 0 since 0 < \tan x < 1 for 0 < x < \frac{\pi}{4} (hence, the integrand is positive everywhere on the interval of integration, so the integral is positive). Thus,

        \[ f(n) - f(n+1) > 0 \quad \implies \quad f(n) > f(n+1). \qquad \blacksquare \]

  2. Proof. We compute, recalling the trig identity \tan^2 x + 1 = \sec^2 x and that the derivative of \tan x is \sec^2 x,

        \begin{align*} f(n) + f(n-2) &= \int_0^{\frac{\pi}{4}} \tan^n x \, dx + \int_0^{\frac{\pi}{4}} \tan^{n-2} x \, dx \\[9pt] &= \int_0^{\frac{\pi}{4}} (\tan^n x + \tan^{n-2} x ) , dx \\[9pt] &= \int_0^{\frac{\pi}{4}} \tan^{n-2} x(\tan^2 x +1) \, dx \\[9pt] &= \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx. \end{align*}

    Now, we make the substitution u = \tan x, du = \sec^2 x \, dx and so,

        \begin{align*} f(n) + f(n-2) &= \int_0^{\frac{\pi}{4}} \tan^{n-2} x \sec^2 x \, dx \\[9pt] &= \int_0^1 u^{n-2} \, du \\ &= \frac{u^{n-1}}{n-1} \Bigr \rvert_0^1 \\[9pt] &= \frac{1}{n-1}. \qquad \blacksquare \end{align*}

  3. Proof. From part (a) we have f(n+1) < f(n) so f(n) < f(n-1) < f(n-2); hence,

        \[ f(n) + f(n-2) > 2f(n). \]

    From part (b) we know f(n) + f(n-2) = \frac{1}{n-1}. Therefore,

        \[ 2f(n) < \frac{1}{n-1}. \]

    For the inequality on the right we again use the inequality in part (a), f(n+2) < f(n+1) < f(n) and so,

        \[ f(n+2) + f(n) < 2f(n). \]

    Using part (b) again we know f(n+2) + f(n) = \frac{1}{(n+2)-1} = \frac{1}{n+1}. Hence,

        \[ \frac{1}{n+1} < 2f(n). \]

    Putting these together we have

        \[ \frac{1}{n+1} < 2f(n) < \frac{1}{n-1}. \qquad \blacksquare \]

One comment

  1. Mihajlo says:

    Comparison theorem, that I think you are using here, has less-than-or-equal relationship, and not strictly less-than. Where are you basing your assumption on?

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