JEE Main 2023 · 1 February · evening · Single correct · +4 / −1

The value of the integral
∫_(- π/4)^(π/4) (x + π/4)/(2 - cos 2x)dx is :

  1. (a) (π ²)/6√3
  2. (b) (π ²)/6
  3. (c) (π ²)/3√3
  4. (d) (π ²)/12√3
Answer and solution

Answer (a) (π ²)/6√3

Solution

Maths sir · 4 steps

Split the numerator: the x part is odd and vanishes, leaving only a constant times an even integral.

  1. Split the integrand. Write it as x/(2-cos 2x) + π/4·1/(2-cos 2x). Since cos 2x is even, x/(2-cos 2x) is odd, so its integral over [-π/4, π/4] is 0.
  2. Use evenness. The second part is even, so I = π/4· 2∫₀^(π/4)dx/(2-cos 2x) = π/2∫₀^(π/4)dx/(2-cos 2x).
  3. Rewrite the denominator. 2-cos 2x = 2(sin² x+cos² x)-(cos² x-sin² x) = 3 sin² x+cos² x. Divide top and bottom by cos² x and put t=tan x, so dt=sec² x dx and t runs from 0 to 1.

    ∫₀^(π/4)(sec² x dx)/(3 tan² x+1) = ∫₀¹dt/(1+3t²) = 1/√3 tan⁻¹(√3 t)|₀¹ = 1/√3·π/3

  4. So. I = π/2·π/3√3 = π²/6√3, which is option (a).

The idea to keep. On a symmetric interval [-a, a], always split the integrand into odd and even parts first. The odd part gives 0, and the even part becomes twice the integral from 0 to a.

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

Study Definite Integration

Asked about 230 times in JEE Main since 2020. This question is from 2023.