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JEE Main 2019 · 9 Jan · Shift II

JEE Main 2019 Mathematics question on Definite Integrals and Areas

If ∫0π/3tan⁡θ2ksec⁡θ dθ=1−12\int_0^{\pi/3} \frac{\tan\theta}{\sqrt{2k\sec\theta}}\, d\theta = 1 - \frac{1}{\sqrt{2}}, (k>0)(k > 0), then the value of kk is
  1. A.
    1
  2. B.
    12\frac{1}{2}
  3. C.
    2
  4. D.
    4
Show answer and solution

Correct option: C

Solution

Rewrite the integrand: tan θ / √(2k sec θ) = (1/√(2k)) · sin θ / √(cos θ).

Integrate: ∫₀^(π/3) sin θ · (cos θ)^(-1/2) dθ. Substitute u = cos θ, du = -sin θ dθ.

Limits: θ=0 → u=1, θ=π/3 → u=1/2.

Integral becomes ∫₁^(1/2) -u^(-1/2) du = [2√u]₁^(1/2) ... wait, = 2√1 - 2√(1/2) = 2 - √2.

So the full expression: (1/√(2k)) · (2 - √2) = 1 - 1/√2 = 1 - √2/2 = (2-√2)/2.

Thus (1/√(2k)) · (2-√2) = (2-√2)/2, giving 1/√(2k) = 1/2, so √(2k) = 2, hence 2k = 4, k = 2.

[JEE Main 2019 · 9 Jan · Shift II]

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