Question

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

• A. $\pi$
• B. $\pi/2$
• C. $\pi/4$
• D. 0

Hint:

For any integral of the form $\int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} \, dx$ or $\int_{0}^{\pi/2} \frac{\cos^n x}{\sin^n x + \cos^n x} \, dx$, the answer is always $\frac{\pi}{4}$, regardless of the value of $n$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a classic problem involving complementary trigonometric functions over the quadrant \([0, \pi/2]\).
Because \(\sin(\pi/2 - x) = \cos x\), any power of sine will transform into the same power of cosine when reflected.
This symmetry ensures that the contribution of the sine term and the cosine term to the total area is identical.
Step 2: Key Formula or Approach:
Using the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\) with \(a = \pi/2\).
Step 3: Detailed Explanation:
Let \( I = \int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx \quad \dots (1) \).
Applying the property \(x \to \pi/2 - x\):
\[ I = \int_{0}^{\pi/2} \frac{\sin^{3/2}(\pi/2 - x)}{\sin^{3/2}(\pi/2 - x) + \cos^{3/2}(\pi/2 - x)} dx \]
Since \(\sin(\pi/2 - x) = \cos x\) and \(\cos(\pi/2 - x) = \sin x\):
\[ I = \int_{0}^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} dx \quad \dots (2) \]
Adding (1) and (2):
\[ 2I = \int_{0}^{\pi/2} \frac{\sin^{3/2} x + \cos^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx = \int_{0}^{\pi/2} 1 dx \]
\[ 2I = [x]_{0}^{\pi/2} = \pi/2 \implies I = \pi/4 \]
Step 4: Final Answer:
The result is \(\pi/4\).