How to evaluate $\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx$ by hand

For the second one,

$$ \begin{align*} \int_{0}^{\infty} \frac{\sin x}{e^x - 1} \; dx &= \int_{0}^{\infty} \frac{\sin x \, e^{-x}}{1 - e^{-x}} \; dx \\ &= \int_{0}^{\infty} \left( \sum_{n=1}^{\infty} \sin x \, e^{-nx} \right) \; dx \\ &\stackrel{\ast}{=} \sum_{n=1}^{\infty} \int_{0}^{\infty} \sin x \, e^{-nx} \; dx \\ &= \sum_{n=1}^{\infty} \frac{1}{1+n^2}, \end{align*}$$

where the starred identity is justified by the following formula

$$ \begin{align*} \int_{0}^{\infty} \frac{\sin x}{e^x - 1} \; dx &= \int_{0}^{\infty} \left( \frac{1 - e^{-Nx}}{1 - e^{-x}} e^{-x} + \frac{e^{-Nx}}{e^x - 1} \right) \sin x \; dx \\ &= \sum_{n=1}^{N} \int_{0}^{\infty} \sin x \, e^{-nx} \; dx + \int_{0}^{\infty} \frac{\sin x \, e^{-Nx}}{e^x - 1} \; dx, \end{align*}$$

together with the dominated convergence theorem. Now the resulting infinite summation can be evaluated in numerous ways. For example, exploiting identities involving the digamma function,

$$ \sum_{n=1}^{\infty} \frac{1}{1+n^2} = \frac{1}{2i} \sum_{n=1}^{\infty} \left( \frac{1}{n-i} - \frac{1}{n+i} \right) = \frac{\psi_0(1+i) - \psi_0(1-i)}{2i} = -\frac{1}{2} + \frac{\pi}{2} \coth \pi. $$

Similar techniques apply to the first integral.

For the first one for example by using the rectangular contour $-R\to +R \to +R+2\pi i \to -R+2\pi i \to -R$
and since there are two simple poles from $\cosh(z)$ at $z=\frac {\pi i}2$ and $z=\frac {3\pi i}2$ of values $-ie^{-\frac {\pi i}2}$ and $ie^{-\frac {3\pi i}2}$ we have : $$2\pi i\ \left(\rm{Res}\left(\frac {e^{iz}}{cosh(z)}; \frac {\pi i}2\right)+\rm{Res}\left(\frac {e^{iz}}{cosh(z)}; \frac {3\pi i}2\right)\right)=2\pi i \left(-ie^{-\frac {\pi}2}+ie^{-\frac {3\pi}2}\right)$$

so that the integral over the rectangular contour of $\dfrac {e^{i x}}{\cosh(x)}$ will be : $$\int_{-R}^R \frac {e^{i x}}{\cosh(x)}\;dx+\int_0^{2\pi} \frac {e^{i R-y}}{\cosh (R+iy)}\;dy+\int_R^{-R} \frac {e^{-2\pi+i x}}{\cosh(2\pi i+x)}\;dx+\int_{2\pi}^0 \frac {e^{-i R-y}}{\cosh (-R+iy)}\;dy=2\pi \left(e^{-\frac {\pi}2}-e^{-\frac {3\pi}2}\right)$$

I'll let you prove that the two integrals in $y$ disappear as $R\to\infty$ so that only remains :

$$\lim_{R\to \infty} \int_{-R}^R \frac {e^{i x}}{\cosh(x)}\;dx-\lim_{R\to \infty}\int_{-R}^R \frac {e^{-2\pi-i x}}{\cosh(2\pi i-x)}\;dx=2\pi \left(e^{-\frac {\pi}2}-e^{-\frac {3\pi}2}\right)$$

it is easy to show that $\ \cosh(2\pi i+x)=\cosh(x)$ (use exponential notation) so that :

$$(1-e^{-2\pi})\int_{-\infty}^\infty \frac {e^{i x}}{\cosh(x)}\;dx=2\pi \left(e^{-\frac {\pi}2}-e^{-\frac {3\pi}2}\right)$$

and the result (keeping the real part) : $$\int_{-\infty}^\infty \frac {\cos x}{\cosh x}\;dx=2\pi \frac{e^{-\frac {\pi}2}}{1+e^{-\pi}}=\frac{\pi}{\cosh \frac{\pi}2}$$

(the second one may be solved the same way...)

Note

\begin{align} I(a)=&\int_{0}^\infty \frac{\cosh ax}{\cosh x}\,dx\\ \overset{t=e^{-x}}=&\int_0^1 \frac{t^a}{1+t^2}dt + \int_0^1 \frac{t^{-a}}{1+t^2}\overset{t\to1/t}{dt}\\ =&\int_0^\infty \frac{t^a}{1+t^2}dt=\frac\pi2\sec\frac{a\pi}2 \end{align} where the last step is well known, e.g. here. Then $$\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,dx = 2I(i)=\pi\>\text{sech}\frac\pi2 $$

Let us start from the Weierstrass product for the sine function: $$ \sin(x) = x \prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right)\tag{1} $$ This identity over the real line can be proved by exploiting Chebyshev polynomials, but it holds over $\mathbb{C}$ and it is equivalent to $$ \sinh(x) = x \prod_{n\geq 1}\left(1+\frac{x^2}{n^2\pi^2}\right).\tag{1'} $$ If we apply $\frac{d}{dx}\log(\cdot)$ (the logarithmic derivative) to both sides of $(1)$ we end up with $$ \cot(x)-\frac{1}{x} = \sum_{n\geq 1}\frac{2x}{x^2-n^2 \pi^2} \tag{2}$$ $$ \coth(x)-\frac{1}{x} = \sum_{n\geq 1}\frac{2x}{x^2 + n^2 \pi^2} \tag{2'}$$ and by expanding the main term of the RHS of $(2)$ as a geometric series we have $$ \frac{1-x\cot x}{2} = \sum_{n\geq 1}\frac{\frac{x^2}{n^2 \pi^2}}{1-\frac{x^2}{n^2\pi^2}} =\sum_{n\geq 1}\sum_{m\geq 1}\left(\frac{x^2}{n^2 \pi^2}\right)^m = \sum_{m\geq 1}x^{2m}\frac{\zeta(2m)}{\pi^{2m}} \tag{3}$$ $$ \frac{1-x\coth x}{2} = \sum_{m\geq 1}(-1)^m x^{2m}\frac{\zeta(2m)}{\pi^{2m}}\tag{3'} $$

$(3')$ alone is enough to prove that $\zeta(2m)$ is always a rational multiple of $\pi^{2m}$, related to a coefficient of the Maclaurin series of $\frac{z}{e^z-1}=-\frac{z}{2}+\frac{z}{2}\coth\frac{z}{2}$, i.e. to a Bernoulli number, but that is not the point here.

Let us deal with the second integral:

$$ \int_{0}^{+\infty}\frac{\sin z}{e^z-1}\,dz = \sum_{m\geq 1}\int_{0}^{+\infty}\sin(z) e^{-mz}\,dz = \sum_{m\geq 1}\frac{1}{m^2+1} $$ can be evaluated through $(2')$ by picking $x=\pi$. In a similar fashion

$$ \int_{0}^{+\infty}\frac{\cos z}{e^z+e^{-z}}\,dz = \sum_{m\geq 0}(-1)^m\int_{0}^{+\infty}\cos(z)e^{-(2m+1)z}\,dz = \sum_{m\geq 0}(-1)^m \frac{(2m+1)}{(2m+1)^2+1} $$

can be computed from the Weierstrass product of the cosine function,

$$ \cos(x) = \prod_{m\geq 0}\left(1-\frac{4x^2}{(2m+1)^2\pi^2}\right)\tag{4} $$

$$ \cosh(x) = \prod_{m\geq 0}\left(1+\frac{4x^2}{(2m+1)^2\pi^2}\right)\tag{4'} $$

leading by logarithmic differentiation to

$$ \sum_{m\geq 0}\frac{x}{x^2+(2m+1)^2} = \frac{\pi}{4}\tanh\left(\frac{\pi x}{2}\right)\tag{5} $$

and also to

$$ \sum_{m\geq 0}\frac{(-1)^m(2m+1)}{x^2+(2m+1)^2} = \frac{\pi}{4}\operatorname{sech}\left(\frac{\pi x}{2}\right).\tag{6}$$