Not as Hard as It Looks

A tempting method of solving the problem might involve using some substitution method or integration by parts. It turns out these ways aren't extremely effective. One way of solving this would be by use of numerical methods, but one must wonder if there exists an algebraic way of approaching the problem, and there does. First consider the following identity, it is proven at the end of this answer. $\left| \int _{ a }^{ b }{ f(x)g(x)dx } \right| \le \sqrt { \left( \int _{ a }^{ b }{ (f(x))^{ 2 }dx } \right) \left( \int _{ a }^{ b }{ (g(x))^{ 2 }dx } \right) }$ For any continuous functions $f$ and $g$ . This can be applied to our function for values within its domain $D=\{\sin 2x\ge 0\}=[n\pi, n\pi + \frac{\pi}{2}]$ for any integer $n$ . In our particular case, we can rewrite the integral expression as $\int _{ 0 }^{ b }{ \sqrt { \sin t}\sqrt{ \cos t }dt }$ using trigonometric identities. Then we can let $f(t)=\sqrt{\sin t}, g(t)=\sqrt{\cos t}$ , so we can find the right hand side, which would be the maximum value. By substitution into the second expression, we have $\left| \int _{ 0 }^{ b }{ \sqrt { \sin t}\sqrt{ \cos t }dt } \right| \le \sqrt{\int _{ 0 }^{ b }{ (\sqrt { \sin t } )^2dt } \int _{ 0 }^{ b }{ (\sqrt { \cos t } )^2dt } }$ $=\sqrt{(-\cos b + 1)(\sin b)}$ $=\sqrt{\sin b - \frac{1}{2} \sin 2b}$ Now we need to consider the given constraint, that $4\sin b = 1 + 2\sin 2b$ . This could be rearranged as $\frac{1}{2} \sin b - \sin 2b = \frac{1}{4}$ , which could be substituted into what we found: $\sqrt{\sin b - \frac{1}{2} \sin 2b}$ $=\sqrt{\frac{1}{4}}$ $\int _{ 0 }^{ b }{ \sqrt { \frac{\sin 2t}{2}}dt }\le \boxed{0.5}$ Therefore, $0.5$ must be the maximum value, which is within the first period of the function $\sqrt{\frac{\sin 2t}{2}}$ , who has an area of around $0.6654$ according to numerical methods. Now, thus far this explanation means nothing if we cannot prove the identity described above. So let us consider one.

Identity: $\left| \int _{ a }^{ b }{ f(x)g(x)dx } \right| \le \sqrt { \left( \int _{ a }^{ b }{ (f(x))^{ 2 }dx } \right) \left( \int _{ a }^{ b }{ (g(x))^{ 2 }dx } \right) }$ for any continuous functions $f$ and $g$ .

Proof:

Consider the function $h(t)=\int _{ a }^{ b }{ (f(x)+tg(x))^{ 2 }dx } \ge 0$ where $f,g \in \{ \text{continuous functions}: [a,b]\rightarrow ℝ; a,b \in ℝ \}$ for any $t\neq 0$ . It can be expanded to $\int _{ a }^{ b }{ (f(x))^{ 2 }dx+ } 2t\int _{ a }^{ b }{ f(x)g(x)dx } +t^2\int _{ a }^{ b }{ (g(x))^{ 2 } } dx \ge 0$ With respect to $t$ , this is just a quadratic that is always greater than or equal to zero. This necessitates that its dicriminant must always be less than or equal to zero. More particularly: $4\left( \int _{ a }^{ b }{ f(x)g(x)dx } \right) ^{ 2 }-4\int _{ a }^{ b }{ (g(x))^{ 2 } } dx\int _{ a }^{ b }{ (f(x))^{ 2 }dx } \le 0$ $\left( \int _{ a }^{ b }{ f(x)g(x)dx } \right) ^{ 2 } \le \int _{ a }^{ b }{ (g(x))^{ 2 } } dx\int _{ a }^{ b }{ (f(x))^{ 2 }dx }$ $\left| \int _{ a }^{ b }{ f(x)g(x)dx } \right| \le \sqrt{\int _{ a }^{ b }{ (g(x))^{ 2 } } dx\int _{ a }^{ b }{ (f(x))^{ 2 }dx }}$ $\text{Q.E.D.}$