If f(θ) = sinθ + ∫ _ - (π)/(2) ^ (π)/(2) (sinθ + tcosθ) · f(t)dt ,...

Given $f\left(\theta\right) = sin\theta + {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\left(sin\theta + tcos\theta\right)f\left(t\right)dt$

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

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

So $f\left(\theta\right) = sin\theta + Asin\theta + Bcos\theta$

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

$A = {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\left[\left(A + 1\right)\sin t + Bcost\right]dt$

$⇒ A = \left(A + 1\right){∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}sintdt + B{∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}costdt$

$⇒ A = 2B \ldots \left(1\right)$

$B = {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}t\left(\left(A + 1\right)\sin t + Bcost\right)dt$

$⇒ B = {∫}_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}t\left(A + 1\right)sintdt$

$⇒ B = \left(A + 1\right)2{∫}_{0}^{\dfrac{\pi}{2}}tsintdt$

$⇒ B = \left(A + 1\right)2$

$⇒ 2A + 2 - B = 0 ⋯ \left(2\right)$

After solving

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

\left|{∫}_{0}^{\dfrac{\pi}{2}}f\left(\theta\right)d\theta\right| = \left|{∫}_{0}^{\dfrac{\pi}{2}}\left(- \dfrac{1}{3}sin\theta - \dfrac{2}{3}cos\theta\right)d\theta\right|$</p><p>$= \left|- \dfrac{1}{3}{∫}_{0}^{\dfrac{\pi}{2}}sin\thetad\theta - \dfrac{2}{3}{∫}_{0}^{\dfrac{\pi}{2}}cos\thetad\theta\right| = 1$