Question

Let $f(\theta )=\sin \theta +\underset{-\pi /2}{\overset{\pi /2}{\int }}(\sin \theta +t\cos \theta )f(t)dt$. Then the value of $\left|\underset{0}{\overset{\pi /2}{\int }}f(\theta )d\theta \right|$ is ___

Answer 1

Answer: 1

$f(\theta )=\sin \theta +\underset{-\pi /2}{\overset{\pi /2}{\int }}(\sin \theta +t\cos \theta )f(t)dt$
$f(\theta )=\sin \theta +\sin \theta \underset{-\pi /2}{\overset{\pi /2}{\int }}f(t)dt+\cos \theta \underset{-\pi /2}{\overset{\pi /2}{\int }}tf(t)dt$
Let $A=\underset{-\pi /2}{\overset{\pi /2}{\int }}f(t)dt,B=\underset{-\pi /2}{\overset{\pi /2}{\int }}tf(t)dt$
$f(\theta )=\sin \theta +A\sin \theta +B\cos \theta$
$f(\theta )=(A+1)\sin \theta +B\cos \theta$
$A=\underset{-\pi /2}{\overset{\pi /2}{\int }}(A+1)\sin t+B\cos tdt$
$A=2B\dots \dots (1)$
$B=\underset{-\pi /2}{\overset{\pi /2}{\int }}t((A+1)\sin t+B\cos t)$
$B=\underset{-\pi /2}{\overset{\pi /2}{\int }}t(A+1)\sin t$
$B=(A+1)2\underset{0}{\overset{\pi /2}{\int }}t\sin tdt$
$B=(A+1)2.1$
$2A+2-B=0\dots \dots (2)$
After solving
$B=-\frac{2}{3},A=-\frac{4}{3}$
$\left|\underset{0}{\overset{\pi /2}{\int }}f(\theta )d\theta \right|=\left|\underset{0}{\overset{\pi /2}{\int }}-\frac{1}{3}\sin \theta -\frac{2}{3}\cos \theta \right|$
$=1$