The value of (e^(- π/4) + ∫₀^(π/4) e⁻ˣtan ⁵⁰xdx)/(∫₀^(π/4) e⁻ˣ(tan ⁴⁹x + tan ⁵¹x)dx) is

Solution

Maths sir · 4 steps

Pull out tan⁴⁹ x in the denominator: the bracket becomes sec² x, and that is the derivative we need.

1. Spot the derivative. In the denominator, tan⁴⁹ x + tan⁵¹ x = tan⁴⁹ x (1 + tan² x) = tan⁴⁹ x sec² x. This is exactly 1/50d/dx(tan⁵⁰ x).
2. Integrate by parts. Take e⁻ˣ as the first function and tan⁴⁹ x sec² x as the second, whose integral is (tan⁵⁰ x)/50. The derivative of e⁻ˣ is -e⁻ˣ, so the minus sign turns into a plus.

   ∫₀^(π/4) e⁻ˣ tan⁴⁹ x sec² x dx = 1/50[e⁻ˣ tan⁵⁰ x]₀^(π/4) + 1/50∫₀^(π/4) e⁻ˣ tan⁵⁰ x dx

3. Boundary term. At x = π/4, tan x = 1, so the term is e^(-π/4). At x = 0, tan x = 0, so the term is 0. The denominator is 1/50(e^(-π/4) + ∫₀^(π/4) e⁻ˣ tan⁵⁰ x dx).
4. So. The bracket is exactly the numerator, so the ratio is numerator/(1/50×numerator) = 50, option (b).

The idea to keep. When you see tanⁿ x + tanⁿ⁺² x, factor it as tanⁿ x sec² x, the derivative of (tanⁿ⁺¹ x)/(n+1). Integrate by parts and the leftover integral usually matches another part of the question.

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