NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice
The integral I = ∫ sin 3 θ cos θ 1 + sin 2 θ 2 d θ simplifies to (where, c is the constant of integration)
Options
- A1 2 ln ⁡ sin ⁡ θ + 1 1 + sin 2 ⁡ θ + c
- B1 2 ln 1 + sin 2 θ + 1 1 + sin 2 θ + c
- Cln sin θ + 1 1 + sin 2 θ + c
- Dln sin 2 θ + 1 + 1 sin 2 θ + 2 + c
Correct answer
B. 1 2 ln 1 + sin 2 θ + 1 1 + sin 2 θ + c
Step-by-step solution
Let 1 + sin 2 ⁡ θ = t ⇒ 2 sin ⁡ θ cos ⁡ θ d θ = d t ∴ I = 1 2 ∫ t - 1 t 2 d t = 1 2 ∫ 1 t - 1 t 2 d t = 1 2 ln ⁡ t + 1 t + c = 1 2 ln ⁡ 1 + sin 2 ⁡ θ + 1 1 + sin 2 ⁡ θ + c