If I_n = int tan^n x dx,and I_0 + I_1 + 2 I_ | vedclass

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(B) The reduction formula for $I_n = \int 	an^n x \ dx$ is given by $I_n + I_{n-2} = \frac{	an^{n-1} x}{n-1}$.Given the expression $I_0 + I_1 + 2 I_

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$5$

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(B) The reduction formula for $I_n = \int 	an^n x \ dx$ is given by $I_n + I_{n-2} = \frac{	an^{n-1} x}{n-1}$.Given the expression $I_0 + I_1 + 2 I_2 + 2 I_3 + 2 I_4 + I_5 + I_6$,we can regroup the terms as follows:$(I_2 + I_0) + (I_3 + I_1) + (I_4 + I_2) + (I_5 + I_3) + (I_6 + I_4)$.Using the reduction formula $I_n + I_{n-2} = \frac{	an^{n-1} x}{n-1}$,we substitute each pair:$(I_2 + I_0) = \frac{	an x}{1}$$(I_3 + I_1) = \frac{	an^2 x}{2}$$(I_4 + I_2) = \frac{	an^3 x}{3}$$(I_5 + I_3) = \frac{	an^4 x}{4}$$(I_6 + I_4) = \frac{	an^5 x}{5}$Summing these,we get $\sum_{K=1}^5 \frac{	an^K x}{K}$.Comparing this with the given sum $\sum_{K=1}^n \frac{	an^K x}{K}$,we find $n = 5$.

If $I_n = \int \tan^n x \ dx$,and $I_0 + I_1 + 2 I_2 + 2 I_3 + 2 I_4 + I_5 + I_6 = \sum_{K=1}^n \frac{\tan^K x}{K}$,then $n = $

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