1. We are given

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

We need to find

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

2. Separate the integral into terms independent of $\theta$:

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

Let

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

Then

$$f(\theta)=\sin\theta+A\sin\theta+B\cos\theta=(1+A)\sin\theta+B\cos\theta.$$

So $f$ must be of the form

$$f(\theta)=\alpha\sin\theta+\beta\cos\theta,$$

with

$$\alpha=1+A,\qquad \beta=B.$$

3. Now compute $A$ and $B$ using this form.

Since

$$f(t)=\alpha\sin t+\beta\cos t,$$

we get

$$A=\int_{-\pi/2}^{\pi/2}(\alpha\sin t+\beta\cos t)dt.$$

Now,

• $\sin t$ is odd, so

$$\int_{-\pi/2}^{\pi/2}\sin t\,dt=0.$$

• $\cos t$ is even, so

$$\int_{-\pi/2}^{\pi/2}\cos t\,dt=2\int_0^{\pi/2}\cos t\,dt=2.$$

Hence

$$A=2\beta.$$

Therefore

$$\alpha=1+A=1+2\beta.$$

4. Next,

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

Now,

• $t\cos t$ is odd $\Rightarrow \displaystyle \int_{-\pi/2}^{\pi/2} t\cos t\,dt=0$.
• $t\sin t$ is even, so

$$\int_{-\pi/2}^{\pi/2} t\sin t\,dt=2\int_0^{\pi/2} t\sin t\,dt.$$

Compute:

$$\int t\sin t\,dt=-t\cos t+\sin t.$$

Thus

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

So

$$\int_{-\pi/2}^{\pi/2} t\sin t\,dt=2.$$

Hence

$$B=2\alpha.$$

But $\beta=B$, so

$$\beta=2\alpha.$$

5. Solve the system:

$$\alpha=1+2\beta, \qquad \beta=2\alpha.$$

Substitute $\beta=2\alpha$ into the first:

$$\alpha=1+4\alpha \Rightarrow -3\alpha=1 \Rightarrow \alpha=-\frac13.$$

Then

$$\beta=2\alpha=-\frac23.$$

Therefore

$$f(\theta)=-\frac13\sin\theta-\frac23\cos\theta.$$

6. Now evaluate

$$\int_0^{\pi/2} f(\theta)\,d\theta =\int_0^{\pi/2}\left(-\frac13\sin\theta-\frac23\cos\theta\right)d\theta.$$

So

$$\int_0^{\pi/2} f(\theta)\,d\theta =-\frac13\int_0^{\pi/2}\sin\theta\,d\theta-\frac23\int_0^{\pi/2}\cos\theta\,d\theta.$$

Both integrals equal $1$, hence

$$\int_0^{\pi/2} f(\theta)\,d\theta=-\frac13-\frac23=-1.$$

Therefore

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

7. Final answer:

$$\boxed{1}$$

Comparison with stored correct answer: the derived answer is $1$, which matches the stored correct answer.