Have I computed this definite integral correctly? $$ \int^\frac{\pi}{6}_0 \tan^38x\sec^38x \ dx
$$ $ u = 8x $ $$ \frac{1}{8} \int^{u =\frac{8\pi}{6}}_{u = 8(0)} \tan^3u\sec^3u \ du \ $$ $$ = \frac{1}{8} \int ^\frac{4\pi}{3}_{0} (\tan u\sec u)^2 \sec u\tan u \ du$$
$ v = \sec u $ $$ = \frac{1}{8} \int ^{v = \sec(\frac{4\pi}{3})}_{v = \sec(0)} v^2(v^2-1) \ dv$$ $$ = \frac{1}{8} [\frac{v^5}{5}-\frac{v^3}{3}]^{-2}_1 $$ $$ = -\frac{9}{20}$$
I'm sorry that I have to ask, but integral sites and my calculator suggest that -9/20 is wrong. Could someone tell me if I've made a mistake in my working
 A: Before computing such an integral, you should first check if it converges.
In your case, the value $x=\dfrac{\pi}{16}\in\left[0;\dfrac{\pi}{6}\right]$ is clearly a problematic point, as
$$\lim_{\underset{x\neq\frac{\pi}{16}}{x\to\frac{\pi}{16}}}(\sec(8x))=+\infty$$
Let $f\colon x\mapsto\tan^3(8x)\sec^3(8x)$ and let us check whether $\displaystyle\int_0^\frac{\pi}{16}f(x)\,\mathrm{d}x$ converges.
In fact, there are many ways to check the latter convergence. I chose to simply compute an antiderivative $F$ of $f$ and then find out its limit at point $x=\dfrac{\pi}{16}$.
We can take for example
$$F\colon x\mapsto-\frac{1}{240}\sec^5(8x)\left(5\cos(16x)-1\right)$$
One can now easily prove that the limit of $F$ when $x$ tends to $\dfrac{\pi}{16}$ is equal to $+\infty$, and it follows that $\displaystyle\int_0^\frac{\pi}{16}f(x)\,\mathrm{d}x$ diverges.
As
$$\int_0^\frac{\pi}{6}f(x)\,\mathrm{d}x=\underbrace{\int_0^\frac{\pi}{16}f(x)\,\mathrm{d}x}_{I_1}+\underbrace{\int_\frac{\pi}{16}^\frac{\pi}{6}f(x)\,\mathrm{d}x}_{I_2}$$
converges iff $I_1$ and $I_2$ converge (by definition), we can finally conclude that
$$\int_0^\frac{\pi}{6}\tan^3(8x)\sec^3(8x)\,\mathrm{d}x$$
does not converge.
