If $$f(\theta) = \sin\theta + \int_{-\pi/2}^{\pi/2} (\sin\theta + t\cos\theta) \cdot f(t)\,dt$$, then $$\left|\int_0^{\pi/2} f(\theta)\,d\theta\right|$$ is ______

We are given $$f( \theta) = \sin \theta + \int_{-\pi/2}^{\pi/2} (\sin \theta + t\cos \theta) \cdot f(t)\,dt$$

We need to find $$\left|\int_0^{\pi/2} f( \theta)\,d \theta \right|$$.

First, we simplify the integral by separating terms that depend on $$ \theta$$ from those that depend on $$t$$:

$$f( \theta) = \sin \theta + \sin \theta \int_{-\pi/2}^{\pi/2} f(t)\,dt + \cos \theta \int_{-\pi/2}^{\pi/2} t \cdot f(t)\,dt$$

Let us define two constants:

$$A = \int_{-\pi/2}^{\pi/2} f(t)\,dt$$ $$-(1)$$

$$B = \int_{-\pi/2}^{\pi/2} t \cdot f(t)\,dt$$ $$-(2)$$

Then $$f( \theta) = (1 + A)\sin \theta + B\cos \theta$$ $$-(3)$$

Finding A: Substituting $$(3)$$ into $$(1)$$:

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

Now, $$\int_{-\pi/2}^{\pi/2} \sin t\,dt = 0$$ (since $$\sin t$$ is an odd function on a symmetric interval).

$$\int_{-\pi/2}^{\pi/2} \cos t\,dt = [\sin t]_{-\pi/2}^{\pi/2} = 1 - (-1) = 2$$

Therefore: $$A = (1+A) \cdot 0 + B \cdot 2 = 2B$$ $$-(4)$$

Finding B: Substituting $$(3)$$ into $$(2)$$:

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

$$B = (1+A)\int_{-\pi/2}^{\pi/2} t\sin t\,dt + B\int_{-\pi/2}^{\pi/2} t\cos t\,dt$$

For $$\int_{-\pi/2}^{\pi/2} t\sin t\,dt$$: since $$t\sin t$$ is an even function (odd $$\times$$ odd = even):

$$= 2\int_0^{\pi/2} t\sin t\,dt = 2\left[-t\cos t + \sin t \right]_0^{\pi/2} = 2(0 + 1 - 0) = 2$$

For $$\int_{-\pi/2}^{\pi/2} t\cos t\,dt$$: since $$t\cos t$$ is an odd function (odd $$\times$$ even = odd):

$$= 0$$

Therefore: $$B = 2(1+A) + 0 = 2(1+A)$$ $$-(5)$$

Solving equations (4) and (5):

From $$(4)$$: $$A = 2B$$. Substituting into $$(5)$$:

$$B = 2(1 + 2B) = 2 + 4B$$

$$-3B = 2$$

$$B = -?\frac{2}{3}$$

From $$(4)$$: $$A = 2 \times \left(-?\frac{2}{3} \right) = -?\frac{4}{3}$$

Substituting back into $$(3)$$:

$$f( \theta) = \left(1 - ?\frac{4}{3} \right)\sin \theta + \left(-?\frac{2}{3} \right)\cos \theta = -?\frac{1}{3}\sin \theta - ?\frac{2}{3}\cos \theta$$

Computing the final integral:

$$\int_0^{\pi/2} f( \theta)\,d \theta = \int_0^{\pi/2} \left(-?\frac{1}{3}\sin \theta - ?\frac{2}{3}\cos \theta \right) d \theta$$

$$= -?\frac{1}{3}[-\cos \theta]_0^{\pi/2} - ?\frac{2}{3}[\sin \theta]_0^{\pi/2}$$

$$= -?\frac{1}{3}(0 - (-1)) - ?\frac{2}{3}(1 - 0)$$

$$= -?\frac{1}{3} - ?\frac{2}{3} = -1$$

Therefore: $$\left|\int_0^{\pi/2} f( \theta)\,d \theta \right| = |-1| = 1$$

The answer is $$1$$.