If f θ = sin θ + ∫ - π 2 π 2 sin θ + t cos θ · f t d t , then…

Given $f\left(\theta \right)=\sin \theta +{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\sin \theta +t\cos \theta \right)f\left(t\right)dt$

$f\left(\theta \right)=\sin \theta +\sin \theta {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\left(t\right)dt+\cos \theta {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}tf\left(t\right)dt$

Let $A={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}f\left(t\right)dt, \mathrm{B}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}tf\left(t\right)dt$

So $f\left(\theta \right)=\sin \theta +A\sin \theta +B\cos \theta$

i.e. $f\left(\theta \right)=\left(A+1\right)\sin \theta +B\cos \theta$

$A={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \left[\left(A+1\right)\sin t+B\cos t\right]dt$

$\Rightarrow A=\left(A+1\right){\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \sin tdt+B{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}} \cos tdt$

$\Rightarrow A=2B \dots \left(1\right)$

$B={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}t\left(\left(A+1\right)\sin t+B\cos t\right)dt$

$\Rightarrow B={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}t\left(A+1\right)\sin tdt$

$\Rightarrow B=\left(A+1\right)2{\int }_0^{\frac{\pi }{2}}t\sin tdt$

$\Rightarrow B=\left(A+1\right)2$

$\Rightarrow 2A+2-B=0 \cdots \left(2\right)$

After solving

$B=-\frac{2}{3}, A=-\frac{4}{3}$

$\left|{\int }_0^{\frac{\pi }{2}}f\left(\theta \right)d\theta \right|=\left|{\int }_0^{\frac{\pi }{2}}\left(-\frac{1}{3}\sin \theta -\frac{2}{3}\cos \theta \right)d\theta \right|$

$=\left|-\frac{1}{3}{\int }_0^{\frac{\pi }{2}}\sin \theta d\theta -\frac{2}{3}{\int }_0^{\frac{\pi }{2}}\cos \theta d\theta \right|=1$