Question

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

• A. \(0\)
• B. \( \dfrac{\pi}{4} \)
• C. \( \dfrac{\pi}{2} \)
• D. \( \pi \)

Hint:

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.

Answer 1

The Correct Option is B

Concept: For definite integrals of the form \[ \int_{0}^{a} f(x)\,dx \] we often use the property \[ \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx \] This symmetry property is very useful for integrals containing trigonometric functions.

Step 1: Let \[ I = \int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x}\,dx \] Using the substitution \(x \rightarrow \frac{\pi}{2}-x\): \[ I = \int_{0}^{\pi/2} \frac{\cos^n x}{\sin^n x + \cos^n x}\,dx \]

Step 2: Add the two integrals. \[ 2I = \int_{0}^{\pi/2} \left( \frac{\sin^n x}{\sin^n x + \cos^n x} + \frac{\cos^n x}{\sin^n x + \cos^n x} \right) dx \] \[ 2I = \int_{0}^{\pi/2} 1\,dx \] \[ 2I = \frac{\pi}{2} \]

Step 3: Compute the value of \(I\). \[ I = \frac{\pi}{4} \]