If An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x = $\pi$/4, then for x 2

Question

If An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x = $\pi$/4, then for x > 2

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An + An −2 =1/n - 1 A_n = Area bounded by the curve " y = \left\{ \tan\left( x \right) \right\}^n = \tan^n \left( x \right) and the lines x = 0, y = 0, and x = $\pi$/4 Therefore, A_n = \int_0^$\pi$/4 \tan^n \left( x \right) d x \Rightarrow A_{n - 2} = \int_0^$\pi$/4 \tan^{n - 2} \left( x \right) d x Consider, " A_n = \int_0^$\pi$/4 \tan^n \left( x \right) d x \Rightarrow A_n = \int_0^$\pi$/4 \left\{ \tan^{n - 2} \left( x \right) \right\}\left\{ \tan^2 \left( x \right) \right\} d x \Rightarrow A_n = \int_0^$\pi$/4 \left\{ \tan^{n - 2} \left( x \right) \right\}\left\{ \sec^2 \left( x \right) - 1 \right\} d x \Rightarrow A_n = \int_0^$\pi$/4 \left\{ \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) - \tan^{n - 2} \left( x \right) \right\} d x \Rightarrow A_n = \int_0^$\pi$/4 \left\{ \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) \right\} d x - \int_0^$\pi$/4 \tan^{n - 2} \left( x \right) d x \Rightarrow A_n + A_{n - 2} = \int_0^$\pi$/4 \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) d x Now, A_n + A_{n - 2} = \int_0^$\pi$/4 \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) d x \text{Let }u = \tan\left( x \right) \Rightarrow d u = \sec^2 \left( x \right)dx Also, when x = 0, u = 0 and when x = $\pi$/4, u = 1 Therefore, A_n + A_{n - 2} = \int_0^$\pi$/4 \tan^{n - 2} \left( x \right) \sec^2 \left( x \right) d x = \int_0^1 \left( u^{n - 2} \right) d u = \left[ \frac{u^{n - 1}}{n - 1} \right]_0^1 = \left[ 1/n - 1 - 0 \right] = 1/n - 1

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