If I_{n}=int_0^{frac{pi}{4}} tan ^n x , dx,th | vedclass

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(B) Given,$I_n = \int_0^{\pi/4} 	an^n x \, dx$. We know that for $I_n = \int_0^{\pi/4} 	an^n x \, dx$,the reduction formula is $I_n + I_{n-2} = \int

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$\frac{1}{12}$

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(B) Given,$I_n = \int_0^{\pi/4} 	an^n x \, dx$. We know that for $I_n = \int_0^{\pi/4} 	an^n x \, dx$,the reduction formula is $I_n + I_{n-2} = \int_0^{\pi/4} 	an^{n-2} x (	an^2 x + 1) \, dx = \int_0^{\pi/4} 	an^{n-2} x \sec^2 x \, dx$. Let $u = 	an x$,then $du = \sec^2 x \, dx$. When $x = 0, u = 0$ and when $x = \pi/4, u = 1$. So,$I_n + I_{n-2} = \int_0^1 u^{n-2} \, du = \left[ \frac{u^{n-1}}{n-1} ight]_0^1 = \frac{1}{n-1}$. For $I_{13} + I_{11}$,we set $n = 13$. Thus,$I_{13} + I_{11} = \frac{1}{13-1} = \frac{1}{12}$.

If $I_{n}=\int_0^{\frac{\pi}{4}} \tan ^n x \, dx$,then $I_{13}+I_{11}=$

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