Maximum and Minimum Value of $f(x)$ 
$$f(x)=\sin(x)+\int_{-\pi/2}^{\pi/2}\left(\sin(x)+t\cos(x)\right)f(t)\,\mathrm dt$$
Find maximum and minimum values of $f(x)$.

I tried to simplify this expression by checking even or odd property of $f(x)$.
We can write the above expression as
$$f(x)=(1+I_1)\sin(x)+I_2\cos(x)$$
where
$$I_1=\int_{-\pi/2}^{\pi/2}f(t)\,\mathrm dt$$
and
$$I_2=\int_{-\pi/2}^{\pi/2}tf(t)\,\mathrm dt$$
if $f$ is even $I_2=0$ and if $f$ is odd $I_1=0$, but if $f$ is even
$$f(x)=(1+I_1)\sin(x)$$ which is odd. Similarly if $f$ is odd
$$f(x)=\sin(x)+I_2\cos(x)$$
which is neither even nor odd.
So $f(x)$ is neither even nor odd. Now to find maxima or minima $f'(x)=0$ i.e.,
$$f'(x)=(1+I_1)\cos(x)-I_2\sin(x)=0$$ $\implies$
$$\tan(x)=\frac{1+I_1}{I_2}$$
Unable to proceed further...
 A: To compute $I_1, I_2$:
We know that $f$ satisfies
$$
(*) f(x) = (1 + I_1) \sin x + I_2 \cos x
$$
We integrate! First integrate both sides of $(*)$ with respect to $x$ over $[-\pi/2,\pi/2]$. The left hand side gives $I_1$. The first term of the right hand side is odd and so vanishes under the integral, and we obtain the relation
$$
I_1 = I_2 \int_{-\pi/2}^{\pi/2} \cos x dx
$$
Second, hit both sides of $(*)$ with a factor of $x$ and integrate over the same interval. $x \cos x$ is odd and it vanishes under the integral; the left hand side gives $I_2$, so we have
$$
I_2 = (1 + I_1) \int_{-\pi/2}^{\pi/2} x \sin x dx
$$
With these two relations, you can compute $I_1, I_2$ with some simple manipulations. Now use your formula for critical points and you'll have found your extrema.
A: Setting
$$
a=\int_{-\pi/2}^{\pi/2}tf(t)dt, b=\int_{\pi/2}^{\pi/2}f(t)dt,
$$
Using the linearity of the integral, we have
$$\tag{1}
f(x)=a\cos x+(1+b)\sin x=\sqrt{a^2+(1+b)^2}\cos(x-\alpha),
$$
where $\alpha$ satisfies
$$
\cos\alpha=\frac{a}{\sqrt{a^2+(1+b)^2}}, \quad \sin\alpha=\frac{1+b}{\sqrt{a^2+(1+b)^2}}
$$
It follows from (1) that
$$\tag{2}
\max f=\sqrt{a^2 +(1+b)^2}, \quad \min f=-\max f.
$$
Multiplying both sides of (1) by $x$ and integrating the resulting identity from $-\pi/2$ to $\pi/2$, we get
\begin{eqnarray}
a
&=&\int_{-\pi/2}^{\pi/2}xf(x)dx=\int_{-\pi/2}^{\pi/2}[ax\cos x+(1+b)x\sin  x]dx\cr
&=&(1+b)\int_{-\pi/2}^{\pi/2}x\sin x dx =2(1+b)\int_0^{\pi/2}x\sin x dx\cr
&=&-2(1+b)x\cos x\Big|_0^{\pi/2}+2(1+b)\int_0^{\pi/2}\cos xdx\cr
&=&2(1+b)\sin x\Big|_0^{\pi/2}=2(1+b),
\end{eqnarray}
t.e.
$$\tag{3}
a=2b+2.
$$
integrating both sides of (1), we get
\begin{eqnarray}
b&=&\int_{-\pi/2}^{\pi/2}f(x)dx=\int_{-\pi/2}^{\pi/2}[a\cos x+(1+b)\sin x]dx\cr
&=&\int_{-\pi/2}^{\pi/2}a\cos xdx=a\sin x\Big|_{-\pi/2}^{\pi/2}\cr
&=&2a
\end{eqnarray}
i.e.
$$\tag{4}
b=2a
$$
Combining (3) and (4) we get
$$
a=-\frac23, \quad b=-\frac43
$$
Hence
$$
f(x)=-\frac23\cos x-\frac13\sin x,
$$
and
$$
\max f=\frac{\sqrt{5}}{3}, \min f=-\frac{\sqrt{5}}{3}.
$$
