JEE Main 2020 · 5 September · evening · Single correct · +4 / −1

If
∫ (cos θ)/(5+7 sin θ-2cos²θ)dθ = Alogₑ|B(θ)|+C,
where C is a constant of integration, then B(θ)/A
can be :

  1. (a) (2 sin θ+1)/5(sin θ+3)
  2. (b) (2 sin θ+1)/(sin θ+3)
  3. (c) 5(2 sin θ+1)/(sin θ+3)
  4. (d) 5(sin θ+3)/(2 sin θ+1)
Answer and solution

Answer (c) 5(2 sin θ+1)/(sin θ+3)

Solution

Maths sir · 4 steps

Put t = sin θ, rewrite cos² θ as 1 - sin² θ, and a quadratic you can factor appears.

  1. Make everything sine. Replace cos² θ by 1 - sin² θ. The denominator becomes 5 + 7 sin θ - 2 + 2 sin² θ = 2 sin² θ + 7 sin θ + 3. Now put t = sin θ, so dt = cos θ dθ.
  2. Factor and split. 2t² + 7t + 3 = (2t+1)(t+3). For partial fractions, put t = -1/2 to get the first numerator 2/5, and t = -3 to get the second numerator -1/5.

    ∫ dt/(2t+1)(t+3) = ∫ ( (2/5)/(2t+1) - (1/5)/(t+3) ) dt = 1/5 ln|(2t+1)/(t+3)| + C

  3. Identify A and B. The factor 2/5 times the 1/2 from integrating 1/(2t+1) gives 1/5. So in its natural form, A = 1/5 and B(θ) = (2 sin θ + 1)/(sin θ + 3).
  4. So. B(θ)/A = 5 × (2 sin θ + 1)/(sin θ + 3) = 5(2 sin θ + 1)/(sin θ + 3), which is option (c).

The idea to keep. When cos θ dθ sits on top and the rest can be written in sin θ, substitute t = sin θ. Then factor the quadratic and use partial fractions.

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

Study Indefinite Integration

Asked about 60 times in JEE Main since 2020. This question is from 2020.