How to evaluate $\int_0^1 {\cos(tx)\over \sqrt{1+x^2}}\,\mathrm{d}x$? I would like to calculate the integral
$$\int_0^1 {\cos(tx)\over \sqrt{1+x^2}}\,\mathrm{d}x$$
I have tried to consider the integral as fourier coefficient of $f(x)=\dfrac1{\sqrt{1+x^2}}$, however no ideas. Besides, let $F(t)$ denote above integral as a function of t, after derivating $F(t)$, I could not make any breakthrough. Does anybody have ideas?
 A: 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
$$tI''-I'-tI=\sqrt2\sin(t).$$
A: 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 $n=2k$: 
$$\boxed{J(t)=\sum\limits_{k=0}^{\infty}\dfrac{(-1)^k}{k!}J_kt^{2k},\quad J_0=\ln(1+\sqrt2),\quad J_k=\dfrac{\sqrt2}{2k}-\dfrac{2k-1}{2k}J_{k-1}}$$
A: $\quad\displaystyle\int_0^\infty{\cos(tx)\over\sqrt{1+x^2}}~dx~=~K_0\Big(|t|\Big),~$ 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$)$.
A: That integral is not a simple question, at all. 
Since the range of integration is bounded to $[0, 1]$ you are allowed to use Series expansion, up to the second (or third) order for the function $\frac{1}{\sqrt{1+x^2}}$.
So, using
$$\frac{1}{\sqrt{1+x^2}} \approx 1 - \frac{x^2}{2} + \frac{3}{8}x^4 + \ldots $$
You can solve easily the integration of 
$$\mathcal{J}(t) = \int_0^1\ \cos(tx)\cdot\left(1 - \frac{x^2}{2}\right)\text{d}x$$
which will lead you to obtain
$$ \dfrac1t\int_0^1 \left(1-\frac {x^2}2\right)\dfrac{d\sin(tx)}{dx}dx = \mathcal{J}(t) = \left.\dfrac1t\left(1-\frac {x^2}2\right)\sin(tx)\right|_{x=0}^{x=1} - \dfrac1t\int_0^1 (-x\sin(tx))dx = \dfrac{\sin t}{2t} - \dfrac1{t^2}\int_0^1 x\dfrac{d\cos(tx)}{dx}dx = \dfrac{\sin t}{2t} - \left.\dfrac{x\cos(tx)}{t^2} \right|_{x=0}^{x=1} - \dfrac1{t^2}\int_0^1 cos(tx)dx = {\dfrac{\sin t}{2t} - \left.\dfrac{\cos x}{t^2} - \dfrac{\sin(tx)}{t^3}\right|_{x=0}^{x=1}} = \left(\dfrac1{2t}-\dfrac1{t^3}\right)\sin t - \dfrac{\cos t}{t^2}.$$
Alternatively, use more terms in the approximation. You don't know the value of $t$, but this means nothing because it's in the cosine function, whilst the square root depends only upon $x$.
To me, it fits.
There are, of course, other methods. Maybe I'll write another one here, later.
A: HINT.-This integral has not an elementary standard expresion. You can developpe in series expansion (at hand or with a calculator) so you get 
$${\cos(tx)\over \sqrt{1+x^2}}= 1+A_1(t)x^2+A_2(t)x^4+A_3(t)x^6+.......$$ then take the values $[...]_0^1$ (After integration of course; you will get a function of $t$).
