If I_n = _ - n ^n ^2 x dx then (where . denotes fractional part f… - Vidyadip

MCQ

If ${I_n} = \int\limits_{ - n}^n {{{\tan }^2}\left\{ x \right\}dx} $ then (where {.} denotes fractional part function and $n \in N$ )

✓
${I_1}{I_2} = 8\left( {{{\sec }^2} 1- 2 - {I_1}} \right)$
B
${I_1}{I_2} = 8\left( {{{\sec }^2}1 - 2 + {I_1}} \right)$
C
${I_1}{I_2} = 8\left( {{{\sec }^2} 1+ 2 - {I_1}} \right)$
D
${I_1}{I_2} = 8\left( {{{\sec }^2} 1+ 2 + {I_1}} \right)$

✓

Answer

Correct option: A.

${I_1}{I_2} = 8\left( {{{\sec }^2} 1- 2 - {I_1}} \right)$

a
$I_{n}=2 n \int_{0}^{1} \tan ^{2} x d x$

$=2 \mathrm{n}(\tan \mathrm{x}-\mathrm{x})_{0}^{1}=2 \mathrm{n}(\tan 1-1)$

$\mathrm{I}_{1} \mathrm{I}_{2}=8(\tan 1-1)^{2}=8\left(\sec ^{2} 1-2 \tan 1\right)$

$=8\left(\sec ^{2} 1-2-I_{1}\right)$

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