Question:medium

Evaluate the definite integral: \( \displaystyle \int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x}\, dx \).

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If the denominator contains \( \sin^n x + \cos^n x \), try the substitution \( x \rightarrow \frac{\pi}{2}-x \). Adding the two resulting integrals simplifies the expression quickly.
Updated On: Apr 17, 2026
  • \(0\)
  • \( \dfrac{\pi}{4} \)
  • \( \dfrac{\pi}{2} \)
  • \( \pi \)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
We are asked to find the value of the definite integral of a specific trigonometric function from 0 to \(\pi/2\).
Step 2: Key Formula or Approach:
This type of integral can be solved efficiently using a property of definite integrals, often called the "King's Property":
\[ \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx \] Here, \(a = \pi/2\).
Step 3: Detailed Explanation:
Let the given integral be \(I\).
\[ I = \int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x}\, dx \quad \dots (1) \] Now, apply the property by replacing \(x\) with \((\pi/2 - x)\):
\[ I = \int_{0}^{\pi/2} \frac{\sin^n (\pi/2 - x)}{\sin^n (\pi/2 - x) + \cos^n (\pi/2 - x)}\, dx \] Using the trigonometric identities \(\sin(\pi/2 - x) = \cos x\) and \(\cos(\pi/2 - x) = \sin x\), the integral becomes:
\[ I = \int_{0}^{\pi/2} \frac{\cos^n x}{\cos^n x + \sin^n x}\, dx \quad \dots (2) \] Now, we add the two expressions for \(I\) (Equation 1 and Equation 2):
\[ I + I = \int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x}\, dx + \int_{0}^{\pi/2} \frac{\cos^n x}{\cos^n x + \sin^n x}\, dx \] Since the limits of integration and the denominators are the same, we can combine the numerators:
\[ 2I = \int_{0}^{\pi/2} \frac{\sin^n x + \cos^n x}{\sin^n x + \cos^n x}\, dx \] The integrand simplifies to 1:
\[ 2I = \int_{0}^{\pi/2} 1\, dx \] Now, evaluate the simple integral:
\[ 2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \] Finally, solve for \(I\):
\[ I = \frac{\pi}{4} \] Step 4: Final Answer:
The value of the definite integral is \(\frac{\pi}{4}\).
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