minimize

The last post got deleted when i was trying to edit it..
btw posting that same stuff again...

f(x) = 1 + ka

where a = \int_{-\pi/2}^{\pi/2}{f(t) sin(x-t)}dt

Now {f(t) sin(x-t)}dt = sin(x-t) + ka sin(x-t)

Integrating both sides we get

\int_{-\pi/2}^{\pi/2}{{f(t) sin(x-t)}dt} = \int_{-\pi/2}^{\pi/2}sin(x-t) + ka \int_{-\pi/2}^{\pi/2}sin(x-t)

a = (1+ ka) \int_{-\pi/2}^{\pi/2}sin(x-t)dt = (1+ ka) \int_{-\pi/2}^{\pi/2}sinxcostdt - cosxsintdt

a = (1+ ka) (sinx\int_{-\pi/2}^{\pi/2}costdt - cosx\int_{-\pi/2}^{\pi/2}sintdt )

a = (1+ ka) 2sinx \Rightarrow cosecx = \frac{2}{a} + 2k \Rightarrow cosecx-2k = \frac{2}{a}

a = \frac{2}{cosecx -2k} \Rightarrow f(x) = 1 + \frac{2k}{cosecx- 2k}

now how to solve it furthe..

2) We have
f(x)=1+k\sin x\int_{-\pi/2}^{\pi/2}f(t)\cos t\ \mathrm dt -k\cos x\int_{-\pi/2}^{\pi/2}f(t)\sin t\ \mathrm dt
That is
f(x)=1+ka\sin x -kb\cos x
where
a=\int_{-\pi/2}^{\pi/2}f(t)\cos t\ \mathrm dt ----(1)
and
b=\int_{-\pi/2}^{\pi/2}f(t)\sin t\ \mathrm dt ---(2)
We then have
\int_0^\pi f(x)\ \mathrm dx=\pi + 2ka ----- (3)
On the other hand from (1), we get
a=\int_{-\pi/2}^{\pi/2}(\cos t+ka\sin t \cos t-kb\cos^2t)\ \mathrm dt=2-\dfrac{bk\pi}{2}
That is
a+\dfrac{bk\pi}{2}=2 ----- (4)
And from (2), we get
b=\int_{-\pi/2}^{\pi/2}(\sin t+ka\sin^2t-kb\cos t\sin t)\ \mathrm dt=\dfrac{ak\pi}{2} ---- (5)
Solving (4) and (5) for a and b, we get
a=\dfrac{2}{1+\left(\frac{k\pi}{2}\right)^2}
and b=\dfrac{k\pi}{1+\left(\frac{k\pi}{2}\right)^2}
Hence
\int_0^\pi f(x)\ \mathrm dx = \pi + \dfrac{8}{\pi}\cdot\dfrac{\frac{k\pi}{2}}{1+\left(\frac{k\pi}{2}\right)^2}
which maximizes when
\dfrac{k\pi}{2}=1\quad\Rightarrow\ \boxed{k=\dfrac{2}{\pi}}

kaymant sir as always, absolutely right [1]

anyone finishing the first one except experts ?

Thanks Anant sir..
I got my mistake...

btw i added 2 more and will keeping adding if previous ones are solved.

1. 5Ï€⁶ 18 ??????

no thats not the ans.
correct ans is \frac{ 315 }{ 4\;\pi^{6}}

thanks sir [1]