If the value of the integral ∫-π/2^(π/2)((x² cos x)/(1+πˣ)+(1+sin ² x)/(1+e^(sin x²¹²³))) d x=π/4(π+a)-2, then the value of a is

Solution

Maths sir · 3 steps

Symmetric limits: add f(x) and f(-x), and both ugly denominators vanish.

1. ∫₋ₐᵃ f(x) dx = ∫₀ᵃ [f(x) + f(-x)] dx

2. First term. Replacing x by -x gives (x²cos x)/(1+π⁻ˣ) = (πˣ x²cos x)/(1+πˣ). Added to the original term, this gives x²cos x. By parts, ∫₀^(π/2) x²cos x dx = [x²sin x + 2x cos x - 2 sin x]₀^(π/2) = π²/4 - 2.
3. Second term. sin²¹²³ x is odd, so replacing x by -x turns eᵗ into e⁻ᵗ. Since 1/(1+eᵗ) + 1/(1+e⁻ᵗ) = 1, the pair adds up to 1+sin² x, and ∫₀^(π/2)(1+sin² x) dx = π/2 + π/4 = 3π/4.
4. So. The integral is π²/4 + 3π/4 - 2 = π/4(π + 3) - 2, so a = 3, option (b).

The idea to keep. When the limits are -a to a and you see 1+kˣ or 1+eᵒᵈᵈ in a denominator, pair f(x) with f(-x). The two fractions add to the numerator alone.

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