Question:
Evaluate : \( \int_{0}^{\pi} \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x} \, dx \)
Evaluate : \( \int_{0}^{\pi} \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x} \, dx \)
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For integrals like \( \int_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} \, dx \), the result is always half the upper limit (\(\pi/4\)).
Be careful with the limits; here the limit was \(\pi\), requiring an extra symmetry step before applying the standard shortcut.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Use the definite integral property \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \).
• Use the property \( \int_0^{2a} f(x) \, dx = 2\int_0^a f(x) \, dx \) if \( f(2a - x) = f(x) \).
• Note that \(\sin(\pi - x) = \sin x\) and \(\cos(\pi - x) = -\cos x\).
Step 1: Test the symmetry of the integrand
Let \(f(x) = \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x}\). \[ f(\pi - x) = \frac{\sin^{2026} (\pi - x)}{\sin^{2026} (\pi - x) + \cos^{2026} (\pi - x)} \] \[ f(\pi - x) = \frac{(\sin x)^{2026}}{(\sin x)^{2026} + (-\cos x)^{2026}} = \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x} = f(x) \] Because \(f(\pi - x) = f(x)\), we can use the property: \[ I = 2 \int_{0}^{\pi/2} \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x} \, dx \quad \dots(1) \]
Step 2: Apply the complementary angle property on the new interval
Let \(I' = \int_{0}^{\pi/2} \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x} \, dx\). Using \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \): \[ I' = \int_{0}^{\pi/2} \frac{\sin^{2026} (\pi/2 - x)}{\sin^{2026} (\pi/2 - x) + \cos^{2026} (\pi/2 - x)} \, dx \] Since \(\sin(\pi/2 - x) = \cos x\) and \(\cos(\pi/2 - x) = \sin x\): \[ I' = \int_{0}^{\pi/2} \frac{\cos^{2026} x}{\cos^{2026} x + \sin^{2026} x} \, dx \quad \dots(2) \]
Step 3: Sum the two forms of the integral
Adding \(I'\) from Step 2 and its original definition: \[ 2I' = \int_{0}^{\pi/2} \frac{\sin^{2026} x + \cos^{2026} x}{\sin^{2026} x + \cos^{2026} x} \, dx \] \[ 2I' = \int_{0}^{\pi/2} 1 \, dx = [x]_0^{\pi/2} = \pi/2 \] \[ I' = \pi/4 \]
Step 4: Calculate final answer \(I\)
From Step 1, \(I = 2I'\): \[ I = 2 \times \frac{\pi}{4} = \frac{\pi}{2} \]
• Use the definite integral property \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \).
• Use the property \( \int_0^{2a} f(x) \, dx = 2\int_0^a f(x) \, dx \) if \( f(2a - x) = f(x) \).
• Note that \(\sin(\pi - x) = \sin x\) and \(\cos(\pi - x) = -\cos x\).
Step 1: Test the symmetry of the integrand
Let \(f(x) = \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x}\). \[ f(\pi - x) = \frac{\sin^{2026} (\pi - x)}{\sin^{2026} (\pi - x) + \cos^{2026} (\pi - x)} \] \[ f(\pi - x) = \frac{(\sin x)^{2026}}{(\sin x)^{2026} + (-\cos x)^{2026}} = \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x} = f(x) \] Because \(f(\pi - x) = f(x)\), we can use the property: \[ I = 2 \int_{0}^{\pi/2} \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x} \, dx \quad \dots(1) \]
Step 2: Apply the complementary angle property on the new interval
Let \(I' = \int_{0}^{\pi/2} \frac{\sin^{2026} x}{\sin^{2026} x + \cos^{2026} x} \, dx\). Using \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \): \[ I' = \int_{0}^{\pi/2} \frac{\sin^{2026} (\pi/2 - x)}{\sin^{2026} (\pi/2 - x) + \cos^{2026} (\pi/2 - x)} \, dx \] Since \(\sin(\pi/2 - x) = \cos x\) and \(\cos(\pi/2 - x) = \sin x\): \[ I' = \int_{0}^{\pi/2} \frac{\cos^{2026} x}{\cos^{2026} x + \sin^{2026} x} \, dx \quad \dots(2) \]
Step 3: Sum the two forms of the integral
Adding \(I'\) from Step 2 and its original definition: \[ 2I' = \int_{0}^{\pi/2} \frac{\sin^{2026} x + \cos^{2026} x}{\sin^{2026} x + \cos^{2026} x} \, dx \] \[ 2I' = \int_{0}^{\pi/2} 1 \, dx = [x]_0^{\pi/2} = \pi/2 \] \[ I' = \pi/4 \]
Step 4: Calculate final answer \(I\)
From Step 1, \(I = 2I'\): \[ I = 2 \times \frac{\pi}{4} = \frac{\pi}{2} \]
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