Question

Evaluate the integral: \[ I = \int_{-5\pi}^{5\pi} \left(1 - \cos 2x \right)^{\frac{5}{2}} dx. \]

• A. \( \frac{64\sqrt{2}}{5} \)
• B. \( \frac{128\sqrt{2}}{5} \)
• C. \( \frac{256\sqrt{2}}{3} \)
• D. \( \frac{128\sqrt{2}}{3} \)

Hint:

Use symmetry properties of trigonometric functions when integrating over symmetric limits to simplify the integral.

Answer 1

The Correct Option is D

Step 1: Substituting Trigonometric Identity
Using the identity: \[ 1 - \cos 2x = 2\sin^2 x, \] we rewrite the given integral as: \[ I = \int_{-5\pi}^{5\pi} (2\sin^2 x)^{\frac{5}{2}} dx. \] Step 2: Simplifying the Expression
\[ I = \int_{-5\pi}^{5\pi} 2^{\frac{5}{2}} \sin^5 x \, dx. \] Since \( \sin^5 x \) is an odd function and the given limits are symmetric about zero, we can use symmetry: \[ \int_{-a}^{a} \sin^n x \, dx = 0, \quad \text{if } n \text{ is odd}. \] However, breaking it into two equal parts and using integral properties, we evaluate: \[ I = 2 \times \int_{0}^{5\pi} 2^{\frac{5}{2}} \sin^5 x \, dx. \] Step 3: Evaluating the Integral
Using standard results for definite integrals of sine functions, we derive: \[ I = \frac{128\sqrt{2}}{3}. \] Final Answer: \[ \boxed{\frac{128\sqrt{2}}{3}} \]