For any integer n geq 2,let I_n = int tan^n x | vedclass

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(A) Given $I_n = \int 	an^n x \, dx$. We can write this as: $I_n = \int 	an^{n-2} x \cdot 	an^2 x \, dx$ $I_n = \int 	an^{n-2} x (\sec^2 x - 1) \,

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$(n-1, 1)$

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(A) Given $I_n = \int 	an^n x \, dx$. We can write this as: $I_n = \int 	an^{n-2} x \cdot 	an^2 x \, dx$ $I_n = \int 	an^{n-2} x (\sec^2 x - 1) \, dx$ $I_n = \int 	an^{n-2} x \sec^2 x \, dx - \int 	an^{n-2} x \, dx$ Using the substitution $u = 	an x$,$du = \sec^2 x \, dx$,the first integral becomes $\int u^{n-2} \, du = \frac{u^{n-1}}{n-1} = \frac{	an^{n-1} x}{n-1}$. Thus,$I_n = \frac{	an^{n-1} x}{n-1} - I_{n-2}$. Comparing this with the given form $I_n = \frac{1}{a} 	an^{n-1} x - b I_{n-2}$,we get: $\frac{1}{a} = \frac{1}{n-1} \implies a = n-1$ $b = 1$ Therefore,the ordered pair $(a, b) = (n-1, 1)$.

For any integer $n \geq 2$,let $I_n = \int \tan^n x \, dx$. If $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$ for $n \geq 2$,then the ordered pair $(a, b)$ is equal to

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