The value of the integral
48/(π ⁴)∫₀^π ( (3π x²)/2 - x³ )(sin x)/(1 + cos ²x)dx is equal to __________.

Solution

Maths sir · 4 steps

Use the King property: replacing x by π - x turns that ugly cubic into a plain constant.

1. Apply King property. Call the integral I and let f(x) = (3π x²)/2 - x³. The factor (sin x)/(1+cos² x) is unchanged when x → π - x, because sin(π - x) = sin x and cos²(π - x) = cos² x. So I is also the integral of f(π - x)(sin x)/(1+cos² x) from 0 to π.
2. Add the two forms. Expanding gives f(π - x) = π³/2 - (3π x²)/2 + x³. Every x term cancels, so f(x) + f(π - x) = π³/2.

   2I = π³/2∫₀^π (sin x)/(1+cos² x) dx = π³/2∫₋₁¹ dt/(1+t²) = π³/2×π/2

3. Evaluate. The substitution t = cos x gives [tan⁻¹ t]₋₁¹ = π/4 + π/4 = π/2. So 2I = π⁴/4 and I = π⁴/8.
4. So. 48/π⁴ × π⁴/8 = 6. The answer is 6.

The common slip: 12. Forgetting that adding the two forms gives 2I, not I, leads to I = π⁴/4 and a final answer of 12.

The idea to keep. When the limits are 0 to π and the trig part stays the same under x → π - x, add I to its King form. A polynomial with the right shape collapses to a constant.

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