For any integer $n \geq 2$, let $I_n=\int \tan ^n x d x$. If $I_n=\frac{1}{a} \tan ^{n-1} x-b I_{n-2}$ for…
For any integer $n \geq 2$, let $I_n=\int \tan ^n x d x$. If $I_n=\frac{1}{a} \tan ^{n-1} x-b I_{n-2}$ for $n \geq 2$, then the ordered pair $(a, b)$ equals to
$\left(n-1, \frac{n-1}{n-2}\right)$
$\left(n-1, \frac{n-2}{n-1}\right)$
$(n, 1)$
$(n-1,1)$
Solution
Given, $I_n=\int \tan ^n x d x$
$
\begin{aligned}
& =\int \tan ^{n-2} x \cdot \tan ^2 x d x \\
& =\int \tan ^{n-2}\left(\sec ^2 x-1\right) d x \\
& =\int \tan ^{n-2} \sec ^2 x d x-\int \tan ^{n-2} d x \\
& =\frac{\tan ^{n-1}}{n-1}-I_{n-2} \\
& \quad\left[\begin{array}{l}
\because \text { put tan } x=t \Rightarrow \sec ^2 x d x=d t \\
\left.\therefore \int \tan ^{n-2} d t=\frac{t^{n-1}}{n-1}=\frac{\tan x^{n-1}}{n-1}\right]
\end{array}\right]
\end{aligned}
$
But it is given
$
\begin{aligned}
& I_n=\frac{1}{a} \tan ^{n-1}-b I_{n-2} \\
\therefore & \quad \frac{1}{a}=\frac{1}{n-1} \text { and } b=1 \\
\Rightarrow \quad & a=n-1, b=1 \\
\therefore \quad & (a, b)=(n-1,1)
\end{aligned}
$