Miscellaneous · Q36: $\int_0^{\pi/4}2\tan^3x\,dx=1-\log 2$

Basics you need

Solution

1. The integrand is continuous on $[0, \frac{\pi}{4}]$

   $\tan x = \frac{\sin x}{\cos x}$ is continuous wherever $\cos x \ne 0$⁠. On $[0, \frac{\pi}{4}]$⁠, $\cos x \ge \frac{1}{\sqrt2} > 0$⁠, so $\tan x$⁠, and with it $2\tan^3 x$⁠, is continuous on the closed interval.

   $\tan x$ blows up at $x = \frac{\pi}{2}$⁠; our interval stops well before that. Continuity on $[a, b]$ is what Theorem 2 of Section 7.8.3 needs.
   Lesson: definite integrals and the fundamental theorem

2. Rewrite $\tan^3 x$

   $$\begin{aligned} 2\tan^3 x &= 2\tan x\cdot\tan^2 x \\ &= 2\tan x\,(\sec^2 x - 1) \\ &= 2\tan x\sec^2 x - 2\tan x . \end{aligned}$$

   So

   $$\begin{aligned} I &= \int_0^{\pi/4} 2\tan x\sec^2 x\,dx \\ &\quad - \int_0^{\pi/4} 2\tan x\,dx . \end{aligned}$$
   This is the same trick as for an odd power of sine (Q35): keep one factor aside and convert the even power with a Pythagorean identity. Here $\sec^2 x$ appears, which is exactly the derivative of $\tan x$⁠, so the first piece becomes a substitution. The split into two integrals uses the linearity of the integral.
   $\tan^2 x = \sec^2 x - 1$⁠. Basics: trigonometric identities

3. First piece: substitute $t = \tan x$

   Put $t = \tan x$⁠, $dt = \sec^2 x\,dx$⁠. At $x = 0$⁠, $t = 0$⁠; at $x = \frac{\pi}{4}$⁠, $t = 1$⁠:

   $$\begin{aligned} \int_0^{\pi/4} 2\tan x\sec^2 x\,dx &= \int_0^1 2t\,dt \\ &= \Big[t^2\Big]_0^1 = 1 . \end{aligned}$$
   The substitution is valid on the whole interval: $\tan x$ is differentiable and increasing on $[0, \frac{\pi}{4}]$ (its derivative $\sec^2 x > 0$⁠), so it maps $[0, \frac{\pi}{4}]$ onto $[0, 1]$ once.
   Section 7.9; power rule. Lesson: integration by substitution

4. Second piece: $\int\tan x\,dx = \log|\sec x|$

   $$\begin{aligned} \int_0^{\pi/4} 2\tan x\,dx &= 2\Big[\log|\sec x|\Big]_0^{\pi/4} \\ &= 2\left(\log\sqrt2 - \log 1\right) \\ &= 2\cdot\tfrac12\log 2 = \log 2 . \end{aligned}$$
   $\sec 0 = 1$⁠, so $\log\sec 0 = 0$⁠; $\sec\frac{\pi}{4} = \sqrt2$⁠, and $\log\sqrt2 = \log 2^{1/2} = \frac12\log 2$⁠. On $[0, \frac{\pi}{4}]$⁠, $\sec x \ge 1 > 0$⁠, so the absolute value bars change nothing.
   Section 7.3.1, (i): $\int\tan x\,dx = \log|\sec x| + C$⁠. Basics: logarithms and 1/x

5. Combine, and check by differentiating

   $$I = 1 - \log 2 \approx 0.306853 .$$

   This is the required result. The antiderivative used is $F(x) = \tan^2 x - 2\log|\sec x|$ (from $t^2$ with $t = \tan x$⁠, and $2\log|\sec x|$⁠). Differentiate:

   $$\begin{aligned} F'(x) &= 2\tan x\sec^2 x - 2\cdot\frac{\sec x\tan x}{\sec x} \\ &= 2\tan x\,(\sec^2 x - 1) = 2\tan^3 x \ ✓ \end{aligned}$$

   valid wherever $\cos x \ne 0$⁠. Directly: $F(\frac{\pi}{4}) - F(0)$ $= (1 - 2\log\sqrt2) - (0 - 0)$ $= 1 - \log 2$ ✓.

   The chain rule on $\log|\sec x|$ gives $\frac{\sec x\tan x}{\sec x} = \tan x$⁠. Another correct antiderivative is $\sec^2 x - 2\log|\sec x|$⁠; it differs from $F$ by the constant 1 (because $\sec^2 x = \tan^2 x + 1$⁠), and the constant cancels in $F(b) - F(a)$⁠. Size check: $2\tan^3 x$ rises from 0 to 2 on an interval of length $\frac{\pi}{4} \approx 0.785$⁠, staying small at first, so a value near 0.31 is reasonable.
   Basics: integration undoes differentiation

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