Evaluating $\int_0^1 \sqrt{1-x^2}\cos(ax) dx$ I want to evaluate this integral
$$I=\int_0^1 \sqrt{1-x^2}\cos(ax) dx \quad (a \in  \mathbb{R})$$
but I cannot find a useful strategy. Could you please give me a hint? 
 A: 
I thought it might be instructive to present an approach that relies on tools acquired in a first course in calculus.  However, we begin with a primer on Bessel functions, a topic not typically introduced in elementary calculus.  




PRIMER $1$:

The Bessel Function, $J_n(a)$, of the first kind and of integer order $n$ can be defined as  
$$J_n(a)=\frac1\pi\int_0^\pi \cos(nx-a\sin(x))\,dx \tag{P1}$$
Then, the zeroth order Bessel function, $J_0(a)$ is expressed as 
$$\begin{align}
J_0(a)&=\frac1\pi\int_0^\pi \cos(a\sin(x))\,dx\\\\
&=\frac1\pi\int_0^{\pi/2} \cos(a\sin(x))\,dx+\frac1\pi\int_{\pi/2}^\pi \cos(a\sin(x))\,dx\tag{P2}\\\\
&=\frac2\pi\int_0^{\pi/2}\cos(a\sin(x))\,dx\tag{P3}
\end{align}$$
where we enforced the substitution $x\mapsto \pi-x$ in the second integral of $(\text{P}2)$ to arrive at $(\text{P}3)$.


PRIMER $2$:

The derivative of $J_0(a)$, is
$$\begin{align}
J_0'(a)&=-\frac1\pi\int_0^\pi \sin(x)\sin(a\sin(x))\,dx\\\\
&=-\frac{1}{2\pi}\left(\int_0^\pi \cos(x-a\sin(x))\,dx-\int_0^\pi \cos(x+a\sin(x))\,dx\right)\tag{P4}\\\\
&=-\frac{1}{2\pi}\left(\int_0^\pi \cos(x-a\sin(x))\,dx+\int_0^\pi \cos(x-a\sin(x))\,dx\right)\tag{P5}\\\\
&=-J_1(a)\tag{P6}
\end{align}$$
where we enforced the substitution $x\mapsto \pi-x$ in the second integral of $(\text{P}4)$ to arrive at $(\text{P}5)$.  


Let $f$ be given by the integral 
$$f(a)=\int_0^1 \sqrt{1-x^2}\,\,(a\cos(tx))\,dx\tag 1$$
Writing $t\cos(xa)$ as $a\cos(xa)=\frac{d\sin(xa)}{dx}$ in $(1)$ reveals
$$f(a)=\int_0^1 \sqrt{1-x^2}\,\frac{d\sin(xa)}{dx}\,dx\tag2$$
Integrating by parts the integral in $(2)$, we obtain
$$f(a)=\int_0^1 \frac{x}{\sqrt{1-x^2}}\sin(xa)\,dx\tag3$$
Enforcing the substation $x\mapsto \sin(x)$ in $(3)$ yields
$$\begin{align}
f(t)&=\int_0^{\pi/2} \sin(x)\sin(a\sin(x))\,dx\\\\
&=-\frac{d}{da}\int_0^{\pi/2} \cos(a\sin(x))\,dx\tag 4\\\\
&=-\frac{d}{da}\left(\frac{\pi}{2}J_0(a)\right)\tag5\\\\
&=\frac\pi2 J_1(a)\tag6
\end{align}$$
In going from $(4)$ to $(5)$, we used $(\text{P}3)$ and in going from $(5)$ to $(6)$, we used $(\text{P}6)$.
Finally, we have
$$\bbox[5px,border:2px solid #C0A000]{\frac{f(a)}{a}=\int_0^1 \sqrt{1-x^2}\cos(ax)\,dx=\frac{\pi}{2}\frac{J_1(a)}a}$$
And we are done!
A: Hint: Maclaurin series in powers of $a$.
EDIT: Alternate hint: Laplace transform.
A: Let 
$$I = \int_0^1 \sqrt{1 - x^2} \cos (ax) \, dx.$$
Integrating by parts we have
$$I = \frac{1}{a} \int_0^1 \frac{x \sin (ax)}{\sqrt{1 - x^2}} \, dx.$$
Now, using a result I show here, namely
$$J_0 (a) = \frac{2}{\pi} \int_0^1 \frac{\cos (ax)}{\sqrt{1 - x^2}} \, dx,$$
where $J_0 (x)$ is the Bessel function of the first kind of order zero, differentiating this result with respect to the parameter $a$ we have
$$J'_0 (a) = - \frac{2}{\pi} \int_0^1 \frac{x \sin (ax)}{\sqrt{1 - x^2}} \, dx.$$
Now from the reasonably well-known result of $J'_0 (x) = - J_1 (x)$ where $J_1(x)$ is the Bessel function of the first kind of order one, we have
$$J_1 (a) = \frac{2}{\pi} \int_0^1 \frac{x \sin (ax)}{\sqrt{1 - x^2}} \, dx,$$
and we conclude
$$I = \frac{\pi}{2a} J_1 (a).$$
