NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice
The integral I = ∫ sin ⁡ 2 θ 1 + cos 2 ⁡ θ 2 sin 2 ⁡ θ d θ simplifies to (where, c is the integration constant)
Options
- Aln sin θ + cos θ + c
- B2 ln sin θ - sin 2 θ 2 + c
- Cln sin θ - sin 2 θ + c
- Dln cos θ + cos 2 θ + c
Correct answer
B. 2 ln sin θ - sin 2 θ 2 + c
Step-by-step solution
I = ∫ cos ⁡ θ 1 + cos 2 ⁡ θ sin ⁡ θ d θ Let sin ⁡ θ = t ⇒ cos ⁡ θ d θ = d t ⇒ I = ∫ 2 - t 2 t d t = 2 ln ⁡ t - t 2 2 + c = 2 ln ⁡ sin ⁡ θ - sin 2 ⁡ θ 2 + c