The value of 8/π∫₀^(π/2) (cos x)²⁰²³/((sin x)²⁰²³ + (cos x)²⁰²³)dx is ___________

Solution

Maths sir · 3 steps

Replace x by π/2 − x: sine and cosine swap, and the two integrals add up to π/2.

1. Name the integral. Let I = ∫₀^(π/2) (cos x)²⁰²³/((sin x)²⁰²³ + (cos x)²⁰²³) dx. The power 2023 is large, but it plays no part in the method.
2. Use the property. By ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a-x) dx with a = π/2, cos x becomes sin x and sin x becomes cos x. So I = ∫₀^(π/2) (sin x)²⁰²³/((cos x)²⁰²³ + (sin x)²⁰²³) dx.

   2I = ∫₀^(π/2) ((cos x)²⁰²³ + (sin x)²⁰²³)/((sin x)²⁰²³ + (cos x)²⁰²³) dx = ∫₀^(π/2) 1 dx = π/2

3. So. I = π/4, and 8/π × π/4 = 2.

The common slip: 4. Stopping at 2I = π/2 and taking I = π/2 gives 4. Divide by 2 before multiplying by 8/π.

The idea to keep. When the integrand is a ratio like g(cos x)/(g(sin x) + g(cos x)) from 0 to π/2, the integral is always half the interval, π/4, whatever the power.

Answer from the NTA final answer key. Solution by PrepWiser.