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If $${I_n} = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^n}x\,dx} $$, then :

JEE · Math · previous-year question

  1. A.$${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$$ are in A.P.correct
  2. B.I2 + I4, I3 + I5, I4 + I6 are in A.P.
  3. C.$${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$$ are in G.P.
  4. D.I2 + I4, (I3 + I5)2, I4 + I6 are in G.P.

Answer

A. $${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$$ are in A.P.

Explanation

$${I_n} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^n}xdx} = \int\limits_{\pi /4}^{\pi /2} {{{\cot }^{n - 2}}x(\cos e{c^2}x - 1)dx} $$ = $$\int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x.co{{\sec }^2}} xdx - \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x} dx$$ $$ = \left. {{{{{\cot }^{n - 1}}x} \over {n - 1}}} \right]_{\pi /4}^{\pi /2} - {I_{n - 2}}$$ $$ = {1 \over {n - 1}} - {I_{n - 2}}$$ $$ \Rightarrow {I_n} + {I_{n - 2}} = {1 \over {n - 1}}$$ $$ \Rightarrow {I_2} + {I_4} = {1 \over 3}$$ $${I_3} + {I_5} = {1 \over 4}$$ $${I_4} + {I_6} = {1 \over 5}$$ $$ \therefore $$ $${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$$ are in A.P.

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