Question

The value of the integral \(\int_{0}^{3\pi/2} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} \, dx\) is:

• A. 0
• B. 1
• C. $\frac{\pi}{4}$
• D. $\frac{3\pi}{4}$

Hint:

For any integral of the form $\int_{0}^{a} \frac{f(x)}{f(x) + f(a-x)} \, dx$, the result is simply half of the upper limit: $\frac{a}{2}$. In this problem, $\frac{3\pi/2}{2} = \frac{3\pi}{4}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The upper limit here is \(3\pi/2\). When applying the property \(x \to (a-x)\), we encounter angles in the third quadrant.
We must carefully evaluate the signs of trigonometric functions in this quadrant to see if the denominator remains invariant.
Step 2: Key Formula or Approach:
Standard property: \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\).
Trigonometric relations:
\( \cos(3\pi/2 - x) = -\sin x \)
\( \sin(3\pi/2 - x) = -\cos x \)
Step 3: Detailed Explanation:
Let \( I = \int_{0}^{3\pi/2} \frac{\cos^{3} x}{\cos^{3} x + \sin^{3} x} dx \quad \dots (1) \).
Replacing \(x\) with \((3\pi/2 - x)\):
\[ I = \int_{0}^{3\pi/2} \frac{(-\sin x)^{3}}{(-\sin x)^{3} + (-\cos x)^{3}} dx \]
The cube of a negative number is negative:
\[ I = \int_{0}^{3\pi/2} \frac{-\sin^{3} x}{-\sin^{3} x - \cos^{3} x} dx = \int_{0}^{3\pi/2} \frac{\sin^{3} x}{\sin^{3} x + \cos^{3} x} dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{3\pi/2} \frac{\cos^{3} x + \sin^{3} x}{\cos^{3} x + \sin^{3} x} dx = \int_{0}^{3\pi/2} 1 dx \]
\[ 2I = [x]_{0}^{3\pi/2} = 3\pi/2 \implies I = 3\pi/4 \]
Step 4: Final Answer:
The value is \(3\pi/4\).