Key Concepts and Formulas

• King's Property of Definite Integrals: For a continuous function $f(x)$ on $[a, b]$, $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$.
• Trigonometric Identities:
  • $\sin(\frac{\pi}{2} - \theta) = \cos \theta$
  • $\sin(\frac{\pi}{2} + \theta) = \cos \theta$
  • $\sin(A-B) + \sin(A+B) = 2 \sin A \cos B$
• Integral Properties:
  • $\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx$
  • Constants can be factored out of integrals.

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Step-by-Step Solution

We are given the equation: $\int\limits_0^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x+\alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2 x) \cos x d x=0 \quad (*)$ Let $I_1 = \int\limits_0^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x$ and $I_2 = \int\limits_0^{\frac{\pi}{4}} f(\cos 2 x) \cos x d x$. The equation is $I_1 + \alpha I_2 = 0$.

Step 1: Splitting the first integral ($I_1$) To unify the limits of integration with $I_2$, we split $I_1$ at $x = \frac{\pi}{4}$. $I_1 = \int\limits_0^{\frac{\pi}{4}} f(\sin 2 x) \sin x d x + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x$ Let $I_{1a} = \int\limits_0^{\frac{\pi}{4}} f(\sin 2 x) \sin x d x$ and $I_{1b} = \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x$. The equation becomes $I_{1a} + I_{1b} + \alpha I_2 = 0$.

Step 2: Transforming $I_{1a}$ using King's Property We apply King's Property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$ to $I_{1a}$ with $a=0$ and $b=\frac{\pi}{4}$. Replace $x$ with $\frac{\pi}{4}-x$. $I_{1a} = \int\limits_0^{\frac{\pi}{4}} f\left(\sin \left(2\left(\frac{\pi}{4}-x\right)\right)\right) \sin\left(\frac{\pi}{4}-x\right) d x$ Simplify the argument of $f$: $\sin\left(2\left(\frac{\pi}{4}-x\right)\right) = \sin\left(\frac{\pi}{2}-2x\right) = \cos 2x$. Thus, $I_{1a} = \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin\left(\frac{\pi}{4}-x\right) d x$

Step 3: Transforming $I_{1b}$ using substitution We want to change the limits of $I_{1b}$ from $[\frac{\pi}{4}, \frac{\pi}{2}]$ to $[0, \frac{\pi}{4}]$. Let $x = \frac{\pi}{2} - u$. Then $dx = -du$. When $x = \frac{\pi}{4}$, $u = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$. When $x = \frac{\pi}{2}$, $u = \frac{\pi}{2} - \frac{\pi}{2} = 0$. Substitute into $I_{1b}$: $I_{1b} = \int\limits_{\frac{\pi}{4}}^{0} f\left(\sin \left(2\left(\frac{\pi}{2}-u\right)\right)\right) \sin\left(\frac{\pi}{2}-u\right) (-du)$ Swap the limits and absorb the negative sign: $I_{1b} = \int\limits_0^{\frac{\pi}{4}} f\left(\sin \left(\pi-2u\right)\right) \sin\left(\frac{\pi}{2}-u\right) du$ Simplify the trigonometric terms: $\sin(\pi-2u) = \sin(2u)$ $\sin(\frac{\pi}{2}-u) = \cos u$ So, $I_{1b} = \int\limits_0^{\frac{\pi}{4}} f(\sin 2u) \cos u \, du$ Replace the dummy variable $u$ with $x$: $I_{1b} = \int\limits_0^{\frac{\pi}{4}} f(\sin 2x) \cos x \, dx$

Step 4: Combining the transformed integrals Substitute the transformed $I_{1a}$ and $I_{1b}$ back into the original equation (from Step 1): $\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin\left(\frac{\pi}{4}-x\right) d x + \int\limits_0^{\frac{\pi}{4}} f(\sin 2x) \cos x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x d x = 0$ This approach seems to have led to a complication as the argument of $f$ is different in the second integral. Let's re-evaluate the transformation of $I_{1b}$.

Alternative Step 3: Transforming $I_{1b}$ using substitution (Revised) Let's use the substitution $x = \frac{\pi}{4} + t$ for $I_{1b} = \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx$. When $x = \frac{\pi}{4}$, $t = 0$. When $x = \frac{\pi}{2}$, $t = \frac{\pi}{4}$. $dx = dt$. $\sin(2x) = \sin\left(2\left(\frac{\pi}{4}+t\right)\right) = \sin\left(\frac{\pi}{2}+2t\right) = \cos 2t$ $\sin x = \sin\left(\frac{\pi}{4}+t\right)$ So, $I_{1b} = \int\limits_0^{\frac{\pi}{4}} f(\cos 2t) \sin\left(\frac{\pi}{4}+t\right) dt$ Replacing $t$ with $x$: $I_{1b} = \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin\left(\frac{\pi}{4}+x\right) dx$

Step 4: Combining the transformed integrals (Revised) Now substitute the transformed $I_{1a}$ and $I_{1b}$ back into the equation $I_{1a} + I_{1b} + \alpha I_2 = 0$: $\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin\left(\frac{\pi}{4}-x\right) d x + \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin\left(\frac{\pi}{4}+x\right) d x + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x d x = 0$ Combine the first two integrals as they have the same integrand argument for $f$ and the same limits: $\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \left[ \sin\left(\frac{\pi}{4}-x\right) + \sin\left(\frac{\pi}{4}+x\right) \right] dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x d x = 0$ Use the sum-to-product identity $\sin(A-B) + \sin(A+B) = 2 \sin A \cos B$ with $A=\frac{\pi}{4}$ and $B=x$: $\sin\left(\frac{\pi}{4}-x\right) + \sin\left(\frac{\pi}{4}+x\right) = 2 \sin\left(\frac{\pi}{4}\right) \cos x = 2 \cdot \frac{1}{\sqrt{2}} \cos x = \sqrt{2} \cos x$ Substitute this back into the equation: $\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \left[ \sqrt{2} \cos x \right] dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x d x = 0$ This can be written as: $\sqrt{2} \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$ Factor out the common integral: $(\sqrt{2} + \alpha) \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$

Step 5: Determining the value of $\alpha$ For this equation to hold true for any continuous function $f$, the coefficient of the integral must be zero, because the integral itself is not necessarily zero. For example, if $f(u)=1$, then $\int_0^{\frac{\pi}{4}} \cos x \, dx = [\sin x]_0^{\frac{\pi}{4}} = \frac{1}{\sqrt{2}} \neq 0$. Therefore, we must have: $\sqrt{2} + \alpha = 0$ Solving for $\alpha$: $\alpha = -\sqrt{2}$

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Common Mistakes & Tips

• Incorrect Substitution: Ensure that when performing substitutions, the limits of integration are correctly transformed and that all occurrences of the original variable are replaced.
• Algebraic Errors in Trigonometric Simplification: Carefully apply trigonometric identities to avoid mistakes in simplifying expressions like $\sin(\frac{\pi}{4}-x) + \sin(\frac{\pi}{4}+x)$.
• Assuming the Integral is Zero: Do not assume that $\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$. This integral depends on the function $f$ and is not generally zero for all continuous functions. The equation must hold for arbitrary $f$.

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Summary The problem was solved by strategically splitting the first integral and applying the King's Property and a suitable substitution to transform its parts. This allowed all terms in the equation to be expressed with the same integration limits and the same functional argument ($f(\cos 2x)$). After combining these transformed integrals and using a trigonometric sum-to-product identity, the equation simplified to a form where a common integral term could be factored out. For the equation to be true for an arbitrary continuous function $f$, the remaining coefficient must be zero, leading to the value of $\alpha$.

The final answer is $\boxed{-\sqrt{2}}$, which corresponds to option (C).