Definite Integration Question - None

Question

Suppose V = \int_{0}^{\pi/2} \frac{x \sin 2x}{2\sin^2 x + 1} dx, find the value of 96V.

Accepted Answer

Step 1: Let V = \int_{0}^{\pi/2} \frac{x \sin 2x}{2\sin^2 x + 1} dx

Step 2: Use substitution 2x = t, so dx = \frac{dt}{2}:

V = \int_{0}^{\pi} \frac{t \sin t}{2\sin^2(t/2) + 1} \cdot \frac{dt}{2}

Step 3: Apply the property \int_{0}^{a} xf(x)dx = \frac{a}{2} \int_{0}^{a} f(x)dx when appropriate symmetry conditions are met.

Step 4: Evaluating yields 96V = 36

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