If ∫_(-π / 2)^(π / 2) (8 √2 cos x d x)/(1+eˢⁱⁿˣ)(1+sin ⁴ x)=α π+β log ₑ(3+2 √2), where α, β are integers, then α²+β² equals :

Solution

Maths sir · 4 steps

Add the integral to its mirror image x → -x, and the awkward 1+eˢⁱⁿˣ disappears.

1. Use x to −x. Call the integral I. Replacing x by -x leaves cos x and sin⁴ x unchanged and turns 1/(1+eˢⁱⁿˣ) into eˢⁱⁿˣ/(1+eˢⁱⁿˣ). Adding the two forms, these two fractions sum to 1.
2. Reduce to t. So 2I = ∫_(-π/2)^(π/2) (8√2 cos x)/(1+sin⁴ x) dx, which has an even integrand, so I = ∫₀^(π/2) (8√2 cos x)/(1+sin⁴ x) dx. Put t = sin x to get I = 8√2∫₀¹ dt/(1+t⁴).

   1/(1+t⁴) = 1/2[(1+1/t²)/((t-1/t)²+2) - (1-1/t²)/((t+1/t)²-2)]

3. Evaluate each part. With u = t - 1/t, running from -∞ to 0, the first part gives 1/√2·π/2 = π/2√2. With v = t + 1/t, running from ∞ to 2, the second part gives 1/2√2 ln(2-√2)/(2+√2) = -ln(3+2√2)/2√2. So ∫₀¹ dt/(1+t⁴) = (π + ln(3+2√2))/4√2.
4. So. I = 8√2 × (π + ln(3+2√2))/4√2 = 2π + 2 ln(3+2√2). Then α = 2, β = 2 and α²+β² = 8.

The common slip: 32. Forgetting to halve after adding I to its mirror image makes the result 4π + 4 ln(3+2√2) and gives 32, not 8.

The idea to keep. On symmetric limits, 1/(1+aᵍ⁽ˣ⁾) with odd g pairs with its mirror image to give exactly 1. For 1/(1+t⁴), split into (1+t²) and (t²-1) over 1+t⁴ and divide by t².

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