Q115 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Mathematics
For $n \geq 2$, let $I_n=\int_0^{\pi / 4} \tan ^n x d x$ and $F_n=I_n+I_{n-2}$. Then, $F_n-F_{n+1}=$
- A$\frac{1}{n}$
- B$\frac{1}{(n-1)}$
- Correct answerC$\frac{1}{n(n-1)}$
- D$1+n$
Solution
$\because I_n=\int_0^{\pi / 4} \tan ^n x d x, n \geq 2$
and
$$
\begin{aligned}
F_n & =I_n+I_{n-2} \\
& =\int_0^{\pi / 4}\left(\tan ^n x+\tan ^{n-2} x\right) d x \\
& =\int_0^{\pi / 4}\left(\tan ^2 x+1\right) \tan ^{n-2} x d x \\
& =\int_0^{\pi / 4} \tan ^{n-2} x \sec ^2 x d x
\end{aligned}
$$
Let $\tan x=t$
$$
\Rightarrow \quad \sec ^2 x d x=d t
$$
Then, $\quad$ at $x=\frac{\pi}{4} \Rightarrow t=1$
and
$$
x=0 \Rightarrow t=0
$$
So,
$$
\begin{aligned}
F_n & =\int_0^1 t^{n-2} d t=\left.\frac{t^{n-1}}{n-1}\right|_0 ^1 \\
& =\frac{1}{n-1}(1)^{n-1}=\frac{1}{n-1} \\
F_{n+1} & =\frac{1}{n}
\end{aligned}
$$
$$
\therefore \quad F_{n+1}=\frac{1}{n}
$$
So, $F_n-F_{n+1}=\frac{1}{n-1}-\frac{1}{n}$
$$
=\frac{n-(n-1)}{n(n-1)}=\frac{1}{n(n-1)}
$$
Hence, the correct answer is option: 'C'.