Let $a,\ b$ are real numbers such that $a+b=1$.

Find the minimum value of the following integral.
\[\int_{0}^{\pi}(a\sin x+b\sin 2x)^{2}\ dx \]
1. We start with the given integral:
   \[
   \int_0^{\pi} (a\sin x + b\sin 2x)^2\ dx
   \]

2. Expand the integrand:
   \[
   (a\sin x + b\sin 2x)^2 = a^2 \sin^2 x + 2ab \sin x \sin 2x + b^2 \sin^2 2x
   \]

3. Split the integral into three parts:
   \[
   \int_0^{\pi} (a\sin x + b\sin 2x)^2\ dx = a^2 \int_0^{\pi} \sin^2 x\ dx + 2ab \int_0^{\pi} \sin x \sin 2x\ dx + b^2 \int_0^{\pi} \sin^2 2x\ dx
   \]

4. Evaluate each integral separately:
   - For \( \int_0^{\pi} \sin^2 x\ dx \):
     \[
     \int_0^{\pi} \sin^2 x\ dx = \int_0^{\pi} \frac{1 - \cos 2x}{2}\ dx = \frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_0^{\pi} = \frac{\pi}{2}
     \]

   - For \( \int_0^{\pi} \sin x \sin 2x\ dx \):
     \[
     \int_0^{\pi} \sin x \sin 2x\ dx = \frac{1}{2} \int_0^{\pi} (\cos x - \cos 3x)\ dx = \frac{1}{2} \left[ \sin x - \frac{\sin 3x}{3} \right]_0^{\pi} = 0
     \]

   - For \( \int_0^{\pi} \sin^2 2x\ dx \):
     \[
     \int_0^{\pi} \sin^2 2x\ dx = \int_0^{\pi} \frac{1 - \cos 4x}{2}\ dx = \frac{1}{2} \left[ x - \frac{\sin 4x}{4} \right]_0^{\pi} = \frac{\pi}{2}
     \]

5. Substitute these results back into the original integral:
   \[
   \int_0^{\pi} (a\sin x + b\sin 2x)^2\ dx = a^2 \cdot \frac{\pi}{2} + 2ab \cdot 0 + b^2 \cdot \frac{\pi}{2} = \frac{\pi}{2} (a^2 + b^2)
   \]

6. Use the condition \(a + b = 1\) to find the minimum value of \(a^2 + b^2\):
   - By the Cauchy-Schwarz inequality:
     \[
     (a + b)^2 \leq 2(a^2 + b^2)
     \]
     Given \(a + b = 1\):
     \[
     1 \leq 2(a^2 + b^2) \implies a^2 + b^2 \geq \frac{1}{2}
     \]

7. The minimum value of \(a^2 + b^2\) is \(\frac{1}{2}\), which occurs when \(a = b = \frac{1}{2}\).

8. Therefore, the minimum value of the integral is:
   \[
   \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}
   \]

The final answer is \(\boxed{\frac{\pi}{4}}\).