Question

\[ \int_0^{\frac{\pi}{2}} \frac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}\,dx = \]

• A. \(\frac{\pi}{2}\)
• B. \(\frac{\pi}{4}\)
• C. \(100\)
• D. \(50\)

Hint:

For definite integrals from \(0\) to \(\frac{\pi}{2}\), use the transformation \(x\to\frac{\pi}{2}-x\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This is a classic definite integral problem that appears complex due to the high exponent (100).
However, it is easily solvable using a symmetry property of definite integrals often referred to as the "King's Rule".
Step 2: Key Formula or Approach:
Property: \( \int_0^a f(x) dx = \int_0^a f(a - x) dx \).
Also, \( \sin(\pi/2 - x) = \cos x \) and \( \cos(\pi/2 - x) = \sin x \).
Step 3: Detailed Explanation:

Let the given integral be \( I \):
\[ I = \int_0^{\pi/2} \frac{\sin^{100} x}{\sin^{100} x + \cos^{100} x} dx \quad \dots \text{(Eq 1)} \]

Applying the property \( \int_0^a f(x) dx = \int_0^a f(a - x) dx \) where \( a = \pi/2 \):
\[ I = \int_0^{\pi/2} \frac{\sin^{100}(\pi/2 - x)}{\sin^{100}(\pi/2 - x) + \cos^{100}(\pi/2 - x)} dx \]

Simplify using trigonometric identities:
\[ I = \int_0^{\pi/2} \frac{\cos^{100} x}{\cos^{100} x + \sin^{100} x} dx \quad \dots \text{(Eq 2)} \]

Now, add Eq 1 and Eq 2:
\[ I + I = \int_0^{\pi/2} \frac{\sin^{100} x}{\sin^{100} x + \cos^{100} x} dx + \int_0^{\pi/2} \frac{\cos^{100} x}{\sin^{100} x + \cos^{100} x} dx \]
\[ 2I = \int_0^{\pi/2} \frac{\sin^{100} x + \cos^{100} x}{\sin^{100} x + \cos^{100} x} dx \]
\[ 2I = \int_0^{\pi/2} (1) dx \]

Integrate:
\[ 2I = [x]_0^{\pi/2} = \pi/2 - 0 = \pi/2 \]

Solve for \( I \):
\[ I = \frac{\pi/2}{2} = \frac{\pi}{4} \]

Step 4: Final Answer:
The value of the definite integral is \( \pi/4 \).