How to evaluate $\int_0^1 {\cos(tx)\over \sqrt{1+x^2}}\,\mathrm{d}x$?

Hint:

Deriving on $t$,

$$I'=-\int_0^1\frac{x\sin(tx)}{\sqrt{1+x^2}}dx.$$

Then by parts,

$$I'=-\left.\sqrt{1+x^2}\sin(tx)\right|_0^1+t\int_0^1\sqrt{1+x^2}\cos(tx)\,dx=-\sqrt2\sin(t)+tJ.$$

Deriving a second time on $t$,

$$I''=-\int_0^1\frac{x^2\cos(tx)}{\sqrt{1+x^2}}dx=-J+I,$$ (with $x^2=1+x^2-1$) so that

$$tI''-I'-tI=\sqrt2\sin(t).$$

Let:

$$E(t)=\int\limits_0^1\dfrac{\exp(jtx)}{\sqrt{1+x^2}}dx,\quad j=\sqrt{-1}$$

Derive it $n$ times with respect to $t$:

$$E^{(n)}(t)=j^n\int\limits_0^1\dfrac{\exp(jtx)}{\sqrt{1+x^2}}x^ndx$$ $$E^{(n)}(0)=j^n\int\limits_0^1\dfrac{x^n}{\sqrt{1+x^2}}dx$$

By parts:

$$I_n=\int\limits_0^1\dfrac{x^n}{\sqrt{1+x^2}}dx=x^{n-1}\left.\sqrt{1+x^2}\right|_0^1-(n-1)\int\limits_0^1\dfrac{1+x^2}{\sqrt{1+x^2}}x^{n-2}dx$$ $$nI_n=\sqrt2-(n-1)I_{n-2}$$
$$I_0=\int\limits_0^1\dfrac{dx}{\sqrt{1+x^2}}=\left.\ln|x+\sqrt{1+x^2}|\right|_0^1 = \ln(1+\sqrt2)$$

Maclaurin series:

$$E(t)=\sum\limits_{n=0}^{\infty}\dfrac1{n!}j^nI_nt^n,\quad J(t)=\int\limits_0^1\dfrac{\cos(tx)}{\sqrt{1+x^2}}dx = \operatorname{Re} E(t)$$

Items with $n=2k-1$ have only imaginary part, so we can take in consideration only items with $n=2k$:

$$\boxed{J(t)=\sum\limits_{k=0}^{\infty}\dfrac{(-1)^k}{k!}J_kt^{2k},\quad J_0=\ln(1+\sqrt2),\quad J_k=\dfrac{\sqrt2}{2k}-\dfrac{2k-1}{2k}J_{k-1}}$$

$\quad\displaystyle\int_0^\infty{\cos(tx)\over\sqrt{1+x^2}}~dx~=~K_0\Big(|t|\Big),~$ see Bessel function for more information. As it stands,

however, the integral cannot be expressed even in terms of such special functions $($unless,

of course, we allow for “incomplete” Bessel functions$)$.