Find $\lambda$ for min. value of $\int_{0}^{\pi}\sin^2(x-{\lambda}x)$

Question

For the minimum value of integral below, find $\lambda$. $$\int_{0}^{\pi}\sin^2(x-{\lambda}x)$$

I tried to solve by $\sin^2(x-{\lambda}x) = \frac{1 - \cos2(x-{\lambda}x)}{2}$ but didn't help so much.

Answer 1

$$\int_{0}^{\pi}\sin^2(x-{\lambda}x)dx=\int_{0}^{\pi}\frac{1 - \cos2(x-{\lambda}x)}{2}dx\\=\frac{\pi}{2}-\frac{\sin2(1-\lambda)x}{4(1-\lambda)}|_{0}^{\pi}\\=\frac{\pi}{2}-\frac{\sin2\pi(1-\lambda)}{4(1-\lambda)}\qquad ,\qquad \lambda\ne 1$$and for $\lambda=1$ the value of integral is $0$.So we can write the above integral as following:$$I=\frac{\pi}{2}(1-sinc(2-2\lambda))$$ therefore for minimizing the integral first we need to maximize $sinc (2-2\lambda)$ which happens in $\lambda=1$ . So the minimum of the integral happens in this point ant its value is $0$.