Common Trig Mistake

By the duplication formulas we have that $\int_{0}^{100}\sin^2(x)dx = 50-\frac{1}{4}\int_{200}^{64\pi}\cos(x)dx$ is between $50+1/5$ and $50+1/4$ , while $\int_{0}^{100}\sin(x^2)dx$ differs very few from the Fresnel integral $\int_{0}^{+\infty}\sin(x^2)dx = \sqrt{\frac{\pi}{8}},$ since: $\int_{100}^{+\infty}\sin(x^2)dx = \frac{1}{2}\int_{10^4}^{+\infty}\frac{\sin x}{\sqrt{x}}dx$ is an integral of an infinitesimal sign-alternating function. The value of this integral is bounded in absolute value by the integral of $\frac{1}{2\sqrt{x}}$ over an interval in $[10^4,+\infty)$ having length $\pi$ , hence it is less than $\frac{\pi}{200}$ . This gives: $3/5 < \int_{0}^{100}\sin(x^2)dx < 2/3,$ $50-\frac{7}{15}<\int_{0}^{100}\sin^2(x)-\sin(x^2)dx<50-\frac{7}{20},$ hence the answer is $50$ .