CUET UG 2025 — Mathematics Statistics & Applications

Let $\displaystyle I = \int_0^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$I=∫0π/2​sin8x+cos8xsin8x​dx ...(1)

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Using the King Property for definite integrals:

$$\int_0^{a} f(x)dx = \int_0^{a} f(a-x)dx$$∫0a​f(x)dx=∫0a​f(a−x)dx

For $a = \frac{\pi}{2}$a=2π​:

$$I = \int_0^{\pi/2} \frac{\sin^8\left(\frac{\pi}{2}-x\right)}{\sin^8\left(\frac{\pi}{2}-x\right) + \cos^8\left(\frac{\pi}{2}-x\right)} dx$$I=∫0π/2​sin8(2π​−x)+cos8(2π​−x)sin8(2π​−x)​dx

Using the complementary angle identities:

$$\sin\left(\frac{\pi}{2}-x\right) = \cos x$$sin(2π​−x)=cosx

$$\cos\left(\frac{\pi}{2}-x\right) = \sin x$$cos(2π​−x)=sinx

This gives:

$\displaystyle I = \int_0^{\pi/2} \frac{\cos^8 x}{\cos^8 x + \sin^8 x} dx$I=∫0π/2​cos8x+sin8xcos8x​dx ...(2)

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Adding equations (1) and (2):

$$I + I = \int_0^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} dx + \int_0^{\pi/2} \frac{\cos^8 x}{\sin^8 x + \cos^8 x} dx$$I+I=∫0π/2​sin8x+cos8xsin8x​dx+∫0π/2​sin8x+cos8xcos8x​dx

$$2I = \int_0^{\pi/2} \frac{\sin^8 x + \cos^8 x}{\sin^8 x + \cos^8 x} dx$$2I=∫0π/2​sin8x+cos8xsin8x+cos8x​dx

$$2I = \int_0^{\pi/2} 1 \, dx$$2I=∫0π/2​1dx

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$$2I = [x]_0^{\pi/2}$$2I=[x]0π/2​

$$2I = \frac{\pi}{2} - 0$$2I=2π​−0

$$2I = \frac{\pi}{2}$$2I=2π​

$$I = \frac{\pi}{4}$$I=4π​

Therefore, $\displaystyle \int_0^{\pi/2} \frac{\sin^8 x}{\sin^8 x + \cos^8 x} dx = \boxed{\frac{\pi}{4}}$∫0π/2​sin8x+cos8xsin8x​dx=4π​​