If I n = ∫ π 4 π 2 cot n x d x , then

If In=π4π2cotnxdx, then
  1. I2+I4,I3+I52,I4+I6 are in G.P.
  2. I2+I4,I3+I5,I4+I6 are in A.P.
  3. 1I2+I4,1I3+I5,1I4+I6 are in A.P.
  4. 1I2+I4,1I3+I5,1I4+I6  are in G.P.

Solution

In=π/4π/2cotnxdx=π/4π/2cotn-2xcosec2x-1dx

=-cotn-1xn-1π/4π/2-In-2

=1n-1-In-2

In+In-2=1n-1

I2+I4=13

I3+I5=14

I4+I6=15

1I2+I4,1I3+I5,1I4+I6  are in A.P.

Asked in: JEE Main 2021 (25 Feb Shift 2)

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