Question

For any integer $n\ge 2$, let $l_n=\int {\tan }^{n}xdx$. If $l_n=\frac{1}{a}{\tan }^{n-1}x-b{l}_{n-2}$for $n\ge 2,$ then the ordered pair $(a,b)=$

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

Answer 1

The correct option is D

$(n-1,1)$

Step1. The given integration :

$l_n=\int {\tan }^{n}xdx$

$$\begin{gathered} =\int {\tan }^{n-2}x{\tan }^{2}xdx \\ =\int {\tan }^{n-2}x(se{c}^{2}x-1)dx \\ =\int ({\tan }^{n-2}xse{c}^{2}x-{\tan }^{n-2}x)dx \end{gathered}$$

Let $\tan x=t$

$\Rightarrow$$se{c}^{2}xdx=dt$

Now, $\int {\tan }^{n-2}xse{c}^{2}xdx=\int t^{n-2}dt$

$=\frac{t^{n-2+1}}{(n-2+1)}$ $\left[\because \int x^{n}dx=\frac{x^{n+1}}{(n+1)}\right]$

$=\frac{t^{n-1}}{(n-1)}$

Now, put the value of $t$ then

$=\frac{{\tan }^{n-1}}{(n-1)}$

Step2. Find the ordered pairs $(a,b)$:

And $l_n=\int {\tan }^{n}xdx$

So, $l_n=\frac{{\tan }^{n-1}x}{n-1}-l_{n-2}$

Comparing with $l_n=\frac{1}{a}{\tan }^{n-1}x-b{l}_{n-2}$then

$a=n-1,b=1$

Hence, the correct option is (D).