If I_n = int_0^{pi / 4} tan^n x , dx,then fra | vedclass

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(D) Given $I_n = \int_0^{\pi / 4} 	an^n x \, dx$. Consider $I_n + I_{n+2} = \int_0^{\pi / 4} 	an^n x (1 + 	an^2 x) \, dx = \int_0^{\pi / 4} 	an^n

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$\frac{1}{I_{11} + I_{13}}$

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(D) Given $I_n = \int_0^{\pi / 4} 	an^n x \, dx$. Consider $I_n + I_{n+2} = \int_0^{\pi / 4} 	an^n x (1 + 	an^2 x) \, dx = \int_0^{\pi / 4} 	an^n x \sec^2 x \, dx$. Let $t = 	an x$,then $dt = \sec^2 x \, dx$. When $x=0, t=0$ and when $x=\pi/4, t=1$. Thus,$I_n + I_{n+2} = \int_0^1 t^n \, dt = \left[ \frac{t^{n+1}}{n+1} ight]_0^1 = \frac{1}{n+1}$. Now,the given expression is $\frac{1}{I_2 + I_4} + \frac{1}{I_3 + I_5} + \frac{1}{I_4 + I_6}$. Using the result $I_n + I_{n+2} = \frac{1}{n+1}$,we have: $I_2 + I_4 = \frac{1}{2+1} = \frac{1}{3} \implies \frac{1}{I_2 + I_4} = 3$. $I_3 + I_5 = \frac{1}{3+1} = \frac{1}{4} \implies \frac{1}{I_3 + I_5} = 4$. $I_4 + I_6 = \frac{1}{4+1} = \frac{1}{5} \implies \frac{1}{I_4 + I_6} = 5$. Sum $= 3 + 4 + 5 = 12$. Checking the options,for $n=11$,$I_{11} + I_{13} = \frac{1}{11+1} = \frac{1}{12}$,so $\frac{1}{I_{11} + I_{13}} = 12$. Therefore,the correct option is $D$.

If $I_n = \int_0^{\pi / 4} \tan^n x \, dx$,then $\frac{1}{I_2 + I_4} + \frac{1}{I_3 + I_5} + \frac{1}{I_4 + I_6} = $

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