AP EAMCET20224 Jul 2022Morning ShiftMathematicsIndefinite IntegrationActual
∫ e cot x sin 2 x 2 log cosec x + sin 2 x d x =
Options
- A- 2 e cot x log cosec 2 x + C
- B- 2 e cot x log cosec x + C
- C- 2 e cot x log cosec x + sin x + C
- D- 2 e cot x log cosec x - cot x + C
Correct answer
B. - 2 e cot x log cosec x + C
Step-by-step solution
I = ∫ e cot x sin 2 x 2 log cosec x + sin 2 x d x = ∫ e cot x cosec 2 x 2 log cosec x + sin 2 x d x Let cot x = t ;   cosec 2 x d x = - d t ⇒ I = - ∫ e t log 1 + t 2 + 2 t 1 + 1 t 2 d t = - ∫ e t log 1 + t 2 + 2 t 1 + t 2 d t = - e t log 1 + t 2 + C = - e cot 2 x log 1 + cot 2 x + C = - 2 e cot 2 x log cosec x + C