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

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

Vedclass · Accepted Answer

$(\frac{1}{5}, 0)$

Vedclass · Answer

(C) Given $I_n = \int 	an^n x dx$. We need to evaluate $I_4 + I_6 = \int 	an^4 x dx + \int 	an^6 x dx$. $I_4 + I_6 = \int (	an^4 x + 	an^6 x) dx$. Factor out $	an^4 x$: $I_4 + I_6 = \int 	an^4 x (1 + 	an^2 x) dx$. Since $1 + 	an^2 x = \sec^2 x$,we have: $I_4 + I_6 = \int 	an^4 x \sec^2 x dx$. Let $u = 	an x$,then $du = \sec^2 x dx$. The integral becomes $\int u^4 du = \frac{u^5}{5} + C$. Substituting back $u = 	an x$,we get: $I_4 + I_6 = \frac{1}{5} 	an^5 x + C$. Comparing this with $a 	an^5 x + b x^5 + C$,we get $a = \frac{1}{5}$ and $b = 0$. Thus,the ordered pair $(a, b) = (\frac{1}{5}, 0)$.

Let $I_n = \int \tan^n x dx, (n > 1)$. If $I_4 + I_6 = a \tan^5 x + b x^5 + C$,where $C$ is the constant of integration,then the ordered pair $(a, b)$ is equal to:

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