Question

For any interger $n\ge 2$, let $I_n=\int ta{n}^{n}xdx$, If $I_n=\frac{1}{a}ta{n}^{n-1}x-b{I}_{n-2}$ for $n\ge 2$, then the ordered pair $(a,b)=$

• A. $(n-1,\frac{n-1}{n-2})$
• B. $(n-1,\frac{n-2}{n-1})$
• C. $(n,1)$
• D. $(n-1,1)$

Answer 1

Answer: (D)

$(n-1,1)$

Given, $I_n=\int {\tan }^{n}xdx=\int {\tan }^{n-2}x{\tan }^{2}xdx$

$=\int {\tan }^{n-2}x({\sec }^{2}x-1)dx=\int {\tan }^{n-2}x{\sec }^{2}xdx-\int {\tan }^{n-2}xdx$

$=\int {\tan }^{n-2}xd\left(\tan x\right)-I_{n-2}$

$=\frac{{\tan }^{n-1}x}{n-1}-I_{n-2}$