AP EAMCET20224 Jul 2022Morning ShiftMathematicsIndefinite IntegrationActual
If I n = ∫ tan n x d x , and I 0 + I 1 + 2 I 2 + 2 I 3 + 2 I 4 + I 5 + I 6 = ∑ K = 1 n tan K x K , then n =
Options
- A6
- B5
- C4
- D3
Correct answer
B. 5
Step-by-step solution
Reduction formulae for I n = ∫ tan n x d x = tan n - 1 x n - 1 - I n - 2 i.e. I n + I n - 2 = tan n - 1 x n - 1 Given I 0 + I 1 + 2 I 2 + 2 I 3 + 2 I 4 + I 5 + I 6 = ∑ K = 1 n tan K x K Now, I 0 + I 1 + 2 I 2 + 2 I 3 + 2 I 4 + I 5 + I 6 = I 2 + I 0 + I 3 + I 1 + I 4 + I 2 + I 5 + I 3 + I 6 + I 4 = tan x 1 + tan 2 x 2 + tan 3 x 3 + tan 4 x 4 + tan 5 x 5 = ∑ K = 1 5 tan K x K Hence, n = 5