If {I_n} = int_{ - n}^n {{{tan }^2}{x}dx} th | vedclass

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(A) Given ${I_n} = \int_{ - n}^n {{{	an }^2}\{x\}dx}$.Since $\{x\}$ is a periodic function with period $1$,we have $\int_{ - n}^n {{{	an }^2}\{x\}dx

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

Vedclass · Answer

(A) Given ${I_n} = \int_{ - n}^n {{{	an }^2}\{x\}dx}$.Since $\{x\}$ is a periodic function with period $1$,we have $\int_{ - n}^n {{{	an }^2}\{x\}dx} = 2n \int_0^1 {{{	an }^2}x dx}$.Evaluating the integral: $\int_0^1 {{{	an }^2}x dx} = \int_0^1 {({{\sec }^2}x - 1)dx} = [	an x - x]_0^1 = 	an 1 - 1$.Thus,${I_n} = 2n(	an 1 - 1)$.For $n=1$,${I_1} = 2(	an 1 - 1)$.For $n=2$,${I_2} = 4(	an 1 - 1) = 2{I_1}$.Then ${I_1}{I_2} = {I_1}(2{I_1}) = 2{I_1}^2 = 2[2(	an 1 - 1)]^2 = 8{(	an 1 - 1)^2}$.Expanding the square: $8({	an ^2}1 - 2	an 1 + 1) = 8({{\sec }^2}1 - 1 - 2	an 1 + 1) = 8({{\sec }^2}1 - 2	an 1)$.Since ${I_1} = 2	an 1 - 2$,we have $2	an 1 = {I_1} + 2$.Substituting this: ${I_1}{I_2} = 8({{\sec }^2}1 - ({I_1} + 2)) = 8({{\sec }^2}1 - 2 - {I_1})$.

If ${I_n} = \int_{ - n}^n {{{\tan }^2}\{x\}dx} $ then (where $\{.\}$ denotes the fractional part function and $n \in N$):

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