calculus - Tough integrals that can be easily beaten by using simple techniques

This question is just idle curiosity. Today I find that an integral problem can be easily evaluated by using simple techniques like my answer to evaluate

\begin{equation}

\int_0^{\pi/2}\frac{\cos{x}}{2-\sin{2x}}dx
\end{equation}

I'm even shocked (and impressed, too) by user @Tunk-Fey's answer and user @David H's answer where they use simple techniques to beat hands down the following tough integrals

\begin{equation}
\int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}\ dx
\end{equation}

and

\begin{equation}
\int_{0}^{\infty}\frac{\ln x}{\sqrt{x}\,\sqrt{x+1}\,\sqrt{2x+1}}\ dx
\end{equation}
So, I'm wondering about super tough definite integrals that can easily beaten by using simple techniques with only a few lines of answer. Can one provide it including its evaluation?

To avoid too many possible answers and to narrow the answer set, I'm interested in knowing tough integrals that can easily beaten by only using clever substitutions, simple algebraic manipulations, trigonometric identities, or the following property

\begin{equation}
\int_b^af(x)\ dx=\int_b^af(a+b-x)\ dx

\end{equation}

I'd request to avoid using contour/ residue integrals, special functions (except gamma, beta, and Riemann zeta function), or complicated theorems. I'd also request to avoid standard integrals like

\begin{align}
\int_{-1}^1\frac{\cos x}{1+e^{1/x}}\ dx&=\sin 1\tag1\\[10pt]
\int_0^\infty\frac{\log ax}{b^2+c^2x^2}\ dx&=\frac{\pi\log\left(\!\frac{ab}{c}\!\right)}{2bc}\tag2\\[10pt]
\int_0^1\frac{1}{(ax+b(1-x))^2}\ dx&=\frac{1}{bc}\tag3\\[10pt]
\int_0^{\pi/2}\frac{\sin^kx}{\sin^kx+\cos^kx}\ dx&=\frac{\pi}{4}\tag4\\[10pt]
\int_0^\infty\frac{e^{-ax}-e^{-bx}}{x}\ dx&=\log\left(\!\frac{b}{a}\!\right)\tag5

\end{align}