Finding an exact solution for u(x)+∫02πcos⁡(x+t)u(t)dt=(π+1)cos⁡x

From
${\displaystyle \cos (a+b)=\cos \left(a\right)\cos \left(b\right)-\sin \left(a\right)\sin \left(b\right)}$
We have that
${\displaystyle u\left(x\right)+\cos \left(x\right){\int }_0^{2\pi }\cos \left(t\right)u\left(t\right)dt-\sin \left(x\right){\int }_0^{2\pi }\sin \left(t\right)u\left(t\right)dt=(\pi +1)\cos \left(t\right)}$
But the integrals are independent of x, so they are constant! So the form of u will be
${\displaystyle u\left(x\right)=(\pi +1)\cos \left(x\right)-c_1\cos \left(x\right)+c_2\sin \left(x\right)}$
And you can calculate the value of the constants easily:
${\displaystyle c_1={\int }_0^{2\pi }\cos \left(t\right)u\left(t\right)dt}$
${\displaystyle c_1={\int }_0^{2\pi }\cos \left(t\right)[(\pi +1)\cos \left(x\right)-c_1\cos \left(x\right)+c_2\sin \left(x\right)]dt}$
${\displaystyle c_1=(\pi +1){\int }_0^{2\pi }{\cos }^2\left(x\right)dx-c_1{\int }_0^{2\pi }{\cos }^2\left(x\right)dx+c_2{\int }_0^{2\pi }\sin \left(x\right)\cos \left(x\right)dx}$
You can easily calculate these integrals with the double-angle formulas:
${\displaystyle c_1=(\pi +1)\pi -c_1\pi +c_{2}0}$
${\displaystyle c_1(1+\pi )=(\pi +1)\pi }$
${\displaystyle c_1=\pi }$
And you can do the same to get ${\displaystyle c_2}$:
${\displaystyle c_2={\int }_0^{2\pi }\sin \left(t\right)u\left(t\right)dt}$
So we have that ${\displaystyle c_1=\pi }$ and ${\displaystyle c_2=0}$, which implies that
${\displaystyle u\left(x\right)=\cos x}$