Let $$f(\theta ) = \sin \theta + \int\limits_{ - \pi /2}^{\pi /2} {(\sin \theta + t\cos \theta )f(t)dt} $$. Then the value of $\left| {\int_0^{\pi /2} {f(\theta )d\theta } } \right|$ is _____________.

$f(\theta)=\sin \theta\left(1+\int_{-\pi / 2}^{\pi / 2} f(t) d t\right)+\cos \theta\left(\int_{-\pi / 2}^{\pi / 2} t f(t) d t\right)$ <br/><br/> Clearly $f(\theta)=a \sin \theta+b \cos \theta$ <br/><br/> Where $a=1+\int_{-\pi / 2}^{\pi / 2}(a \sin t+b \cos t) d t \Rightarrow a=1+2 b\quad\quad...(i)$ <br/><br/> and $b=\int_{-\pi / 2}^{\pi / 2}(a t \sin t+b t \cos t) d t \Rightarrow b=2 a\quad\quad...(ii)$ <br/><br/> from (i) and (ii) we get <br/><br/> $a=-\frac{1}{3} \text { and } b=-\frac{2}{3}$ <br/><br/> So $f(\theta)=-\frac{1}{3}(\sin \theta+2 \cos \theta)$ <br/><br/> $$ \Rightarrow\left|\int_{0}^{\pi / 2} f(\theta) d \theta\right|=\frac{1}{3}(1+2 \times 1)=1 $$