Evalute $\int_0^1 \sqrt{1-x^2} \cos (tx) \, dx$ I can't solve this problem. Can anyone help me?
Let $t \in \mathbb{R}$. Evaluate $\int_0^1 \sqrt{1-x^2} \cos (tx) \, dx.$
Thank you very much!
 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, $J_n(t)$, of the first kind and of integer order $n$ can be defined as  
$$J_n(t)=\frac1\pi\int_0^\pi \cos(nx-t\sin(x))\,dx \tag{P1}$$
Then, the zeroth order Bessel function, $J_0(t)$ is expressed as 
$$\begin{align}
J_0(t)&=\frac1\pi\int_0^\pi \cos(t\sin(x))\,dx\\\\
&=\frac1\pi\int_0^{\pi/2} \cos(t\sin(x))\,dx+\frac1\pi\int_{\pi/2}^\pi \cos(t\sin(x))\,dx\tag{P2}\\\\
&=\frac2\pi\int_0^{\pi/2}\cos(t\sin(x))\,dx\tag{P3}
\end{align}$$
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 $J_0(t)$, is
$$\begin{align}
J_0'(t)&=-\frac1\pi\int_0^\pi \sin(x)\sin(t\sin(x))\,dx\\\\
&=-\frac{1}{2\pi}\left(\int_0^\pi \cos(x-t\sin(x))\,dx-\int_0^\pi \cos(x+t\sin(x))\,dx\right)\tag{P4}\\\\
&=-\frac{1}{2\pi}\left(\int_0^\pi \cos(x-t\sin(x))\,dx+\int_0^\pi \cos(x-t\sin(x))\,dx\right)\tag{P5}\\\\
&=-J_1(t)\tag{P6}
\end{align}$$
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 
$$f(t)=\int_0^1 \sqrt{1-x^2}\,\,(t\cos(tx))\,dx\tag 1$$
Writing $t\cos(xt)$ as $t\cos(xt)=\frac{d\sin(xt)}{dx}$ in $(1)$ reveals
$$f(t)=\int_0^1 \sqrt{1-x^2}\,\frac{d\sin(xt)}{dx}\,dx\tag2$$
Integrating by parts the integral in $(2)$, we obtain
$$f(t)=\int_0^1 \frac{x}{\sqrt{1-x^2}}\sin(xt)\,dx\tag3$$
Enforcing the substation $x\mapsto \sin(x)$ in $(3)$ yields
$$\begin{align}
f(t)&=\int_0^{\pi/2} \sin(x)\sin(t\sin(x))\,dx\\\\
&=-\frac{d}{dt}\int_0^{\pi/2} \cos(t\sin(x))\,dx\tag 4\\\\
&=-\frac{d}{dt}\left(\frac{\pi}{2}J_0(t)\right)\tag5\\\\
&=\frac\pi2 J_1(t)\tag6
\end{align}$$
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
$$\bbox[5px,border:2px solid #C0A000]{\frac{f(t)}{t}=\int_0^1 \sqrt{1-x^2}\cos(tx)\,dx=\frac{\pi}{2}\frac{J_1(t)}t}$$
And we are done!
