If ₀^{π/2} ¹4}≤ft(x/2) \,dx = 2≤ft[ₙ=1}^{7} f(n) - π/4], then f(n)=

Correct Answer is : (−1)n+12n−1{(-1)ⁿ+1}}{2n-1}2n−1(−1)n+1

Correct Answer is : (−1)n+12n−1{(-1)ⁿ+1}}{2n-1}2n−1(−1)n+1

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TG EAMCET 4 May 2025 Shift 2 Paper

Question : 78 of 160

Marks: +1, -0

\int\limits_{0}^{\pi/2} \tan^{14}\left(\frac{x}{2}\right) \,dx = 2\left[\sum\limits_{n=1}^{7} f(n) - \frac{\pi}{4}\right]

f(n)=

Solution:

\text{ Let } I_n = \int\limits_{0}^{\frac{\pi}{2}} \tan^n\left(\frac{x}{2}\right) \,dx

I_{n-2} = \int\limits_{0}^{\pi/2} \left(\tan\frac{x}{2}\right)^{n-2} \,dx

\Rightarrow I_n + I_{n-2} = \int\limits_{0}^{\frac{\pi}{2}} \left(\tan\frac{x}{2}\right)^{n-2} \times \sec^2\frac{x}{2} \,dx

\Rightarrow I_n + I_{n-2} = 2\int\limits_{0}^{1} t^{n-2} \,dt = \frac{2}{n-1}

\Rightarrow I_{14} + I_{12} = \frac{2}{13}

\Rightarrow I_{12} + I_{10} = \frac{2}{11}

\Rightarrow I_{10} + I_8 = \frac{2}{9}

\Rightarrow I_8 + I_6 = \frac{2}{7}

\Rightarrow I_6 + I_4 = \frac{2}{5}

\Rightarrow I_4 + I_2 = \frac{2}{3}

\Rightarrow I_2 + I_0 = 2

\therefore (i)-(ii)+(iii)-(iv)+(v)-(vi)+(vii)

I_{14} = 2\left[\sum\limits_{n=1}^{7} f(x) - \frac{\pi}{4}\right]

\therefore f(n) = \frac{1}{13} - \frac{1}{11} + \frac{1}{9} - \frac{1}{7} \ldots (iv)

f(n) = \frac{(-1)^{n+1}}{2n-1}

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