JEE Main202125 Feb 2021Evening ShiftMathematicsDefinite IntegrationActual
If I n = ∫ π 4 π 2 cot n x d x , then
Options
- AI 2 + I 4 , I 3 + I 5 2 , I 4 + I 6 are in G . P .
- BI 2 + I 4 , I 3 + I 5 , I 4 + I 6 are in A . P .
- C1 I 2 + I 4 , 1 I 3 + I 5 , 1 I 4 + I 6 are in A . P .
- D1 I 2 + I 4 , 1 I 3 + I 5 , 1 I 4 + I 6 are in G . P .
Correct answer
C. 1 I 2 + I 4 , 1 I 3 + I 5 , 1 I 4 + I 6 are in A . P .
Step-by-step solution
I n = ∫ π / 4 π / 2 cot n x d x = ∫ π / 4 π / 2 cot n - 2 x cosec 2 x - 1 d x = - cot n - 1 x n - 1 π / 4 π / 2 - I n - 2 = 1 n - 1 - I n - 2 ⇒ I n + I n - 2 = 1 n - 1 ⇒ I 2 + I 4 = 1 3 I 3 + I 5 = 1 4 I 4 + I 6 = 1 5 ∴ 1 I 2 + I 4 , 1 I 3 + I 5 , 1 I 4 + I 6 are in A . P .