To evaluate the integral, we find the antiderivative of sin (1)/(2) x, which is -2 cos (1)/(2) x. Then, we apply the limits of integration from 0 to (π)/(2).

Substitute the values of cos (π)/(4) = (1)/(√2) and cos 0 = 1 into the expression. Simplify the result to find the value of I.

We use the property of definite integrals that ∫_0^a f(x) dx = ∫_0^a f(a-x) dx. Here, a = 2I. This property is often useful when the integrand has symmetry.

The value of 2I = 2(2-√2) ≈ 1.172. This is not a standard angle like π or (π)/(2). For the property sin(a-x) = sin x and cos(a-x) = -cos x to hold, a must be π. If we assume the problem intended 2I = π, then sin(π-x) = sin x and cos(π-x) = -cos x. Substituting these into the integral, we get a new expression for J.

The value of 2I = 2(2-√2) ≈ 1.172. This is not a standard angle like π or (π)/(2). For the property sin(a-x) = sin x and cos(a-x) = -cos x to hold, a must be π. If we assume the problem intended 2I = π, then sin(π-x) = sin x and cos(π-x) = -cos x. Substituting these into the integral, we get a new expression for J.

If $I = \int_0^{\frac{\pi}{2}} \frac{\sin \frac{1}{2} x}{\sin^2 x + \cos^2 x} \, dx$ , then $\int_0^{2I} \frac{x \sin x \cos x}{\sin^2 x + \cos^4 x} \, dx$ equals:

Answer: The problem statement is mathematically underdetermined due to an inconsistency in the upper limit of the second integral. If $2I$ is assumed to be $\pi$ , the integral can be simplified further, but with $I = 2 - \sqrt{2}$ , $2I \neq \pi$ . Therefore, a unique numerical answer cannot be provided without further clarification or correction of the problem statement.

Step-by-step solution

Step 1: Simplify the first integral

The denominator of the first integral is $\sin^2 x + \cos^2 x$ . This is a fundamental trigonometric identity, which simplifies to $1$ . So, the integral $I$ becomes $\int_0^{\frac{\pi}{2}} \sin \frac{1}{2} x \, dx$ .

Step 2: Evaluate the first integral

To evaluate the integral, we find the antiderivative of $\sin \frac{1}{2} x$ , which is $-2 \cos \frac{1}{2} x$ . Then, we apply the limits of integration from $0$ to $\frac{\pi}{2}$ .

Step 3: Calculate the value of I

Substitute the values of $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos 0 = 1$ into the expression. Simplify the result to find the value of $I$ .

Step 4: Define the second integral and apply property

Let the second integral be $J$ . We substitute the value of $I$ we just calculated into the upper limit of the integral. The integral is of the form $\int_0^a f(x) \, dx$ .

Step 5: Apply the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$

We use the property of definite integrals that $\int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx$ . Here, $a = 2I$ . This property is often useful when the integrand has symmetry.

Step 6: Observe the symmetry of the integrand

The value of $2I = 2(2-\sqrt{2}) \approx 1.172$ . This is not a standard angle like $\pi$ or $\frac{\pi}{2}$ . For the property $\sin(a-x) = \sin x$ and $\cos(a-x) = -\cos x$ to hold, $a$ must be $\pi$ . If we assume the problem intended $2I = \pi$ , then $\sin(\pi-x) = \sin x$ and $\cos(\pi-x) = -\cos x$ . Substituting these into the integral, we get a new expression for $J$ .

Step 7: Add the original and modified integrals (assuming $2I = \pi$ )

Adding the original integral $J$ and the modified integral $J$ (after applying the property), we combine the numerators. This step is part of a common technique to solve integrals of this form, but it requires the upper limit to be $\pi$ for the trigonometric identities to simplify the expression further.

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