Find $\int_0^{\pi/4} \tan^3x \,\mathrm dx$

Question

How can one evaluate $\int_0^{\pi/4} \tan^3x \,\mathrm dx$ ?

I tried using the following reduction formula.

$$\int \tan^3x \,\mathrm dx = \frac{1}{2}\tan^2x - \int \tan x \,\mathrm dx = \frac{1}{2}\tan^2x - \ln|\sec x|$$

But the range of $\sec x$ is $(-\infty,-1] \cup [1, \infty)$

Answer 1

If you make $x=\tan^{-1}(t)$ $$I=\int_0^{\frac \pi 4} \tan^3(x)\,dx=\int_0^1 \frac {t^3}{1+t^2}\,dt$$ Now $t=\sqrt u$ $$I=\frac 12 \int_0^1 \frac u{1+u}\,du=\frac 12 \int_0^1 \frac {1+u-1}{1+u}\,du=\frac 12 \int_0^1\,du-\frac 12 \int_0^1 \frac {1}{1+u}\,du$$ $$I=\frac{1}{2} (1-\log (2))$$

Answer 2

The range of $\sec x$ is not an issue. The antiderivative can be written as $$\int \tan^3 x \, dx = \frac{1}{2} \tan^2 x - \log |\sec x| + C,$$ which is well-defined on the interval $x \in [0, \pi/4]$. Surely there can be no doubt about the first term. The second term, $\log |\sec x|$, is defined since on $x \in [0,\pi/4]$, $\sec x \in [1, \sqrt{2}]$. Then $\log \sec x$ on this interval has range $[0, \log \sqrt{2}]$.