If I_n = int tan^n x , dx (n 1),then I_4 + | vedclass

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(A) Given $I_n = \int 	an^n x \, dx$. We need to find $I_4 + I_6 = \int 	an^4 x \, dx + \int 	an^6 x \, dx$. $I_4 + I_6 = \int (	an^4 x + 	an^6 x

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$\frac{1}{5} 	an^5 x + C$

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(A) Given $I_n = \int 	an^n x \, dx$. We need to find $I_4 + I_6 = \int 	an^4 x \, dx + \int 	an^6 x \, dx$. $I_4 + I_6 = \int (	an^4 x + 	an^6 x) \, dx$. Factor out $	an^4 x$: $I_4 + I_6 = \int 	an^4 x (1 + 	an^2 x) \, dx$. Using the identity $1 + 	an^2 x = \sec^2 x$: $I_4 + I_6 = \int 	an^4 x \sec^2 x \, dx$. Let $u = 	an x$,then $du = \sec^2 x \, dx$. Substituting these into the integral: $I_4 + I_6 = \int u^4 \, du = \frac{u^5}{5} + C$. Substituting back $u = 	an x$: $I_4 + I_6 = \frac{1}{5} 	an^5 x + C$.

If $I_n = \int \tan^n x \, dx$ $(n > 1)$,then $I_4 + I_6 =$

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