NTA Abhyas JEE Main2020MathematicsIndefinite IntegrationPractice
If I = ∫ sin ⁡ x 3 sin ⁡ x + cos ⁡ x + 2 d x and J = ∫ cos ⁡ x 3 sin ⁡ x + cos ⁡ x + 2 d x , then 3 J - I is equal to (where C is the constant of integration)
Options
- Ax + ln 3 sin x + cos x + 2 + c
- Bx - ln 3 sin x + cos x + 2 + c
- Cln 3 sin x + cos x + 2 + c
- D2 x + ln 3 sin x + cos x + 2 + c
Correct answer
C. ln 3 sin x + cos x + 2 + c
Step-by-step solution
3 J - I = ∫ 3 c o s x - s i n x 3 s i n x + c o s x + 2 d x Let 3 s i n x + c o s x + 2 = t ⇒ 3 cos ⁡ x - sin ⁡ x d x = d t So, 3 J - I = ∫ d t t = ln ⁡ t + C Hence, 3 J - I = ln ⁡ 3 sin ⁡ x + cos ⁡ x + 2 + c