JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 12)
If $${I_n} = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^n}x\,dx} $$, then :
$${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$$ are in A.P.
I2 + I4, I3 + I5, I4 + I6 are in A.P.
$${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$$ are in G.P.
I2 + I4, (I3 + I5)2, I4 + I6 are in G.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.
= $$\int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x.co{{\sec }^2}} xdx - \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^{n - 2}}x} dx$$
$$ = \left. {{{{{\cot }^{n - 1}}x} \over {n - 1}}} \right]_{\pi /4}^{\pi /2} - {I_{n - 2}}$$
$$ = {1 \over {n - 1}} - {I_{n - 2}}$$
$$ \Rightarrow {I_n} + {I_{n - 2}} = {1 \over {n - 1}}$$
$$ \Rightarrow {I_2} + {I_4} = {1 \over 3}$$
$${I_3} + {I_5} = {1 \over 4}$$
$${I_4} + {I_6} = {1 \over 5}$$
$$ \therefore $$ $${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$$ are in A.P.
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