Q92 / 150 · MHT CET 2026 (20 April Shift 2) · Mathematics
If $I_n = \int\limits_0^{\pi/4}\tan^n x\ dx, n \in N$ then $I_{n+2} + I_n$ is equal to
- A$\dfrac{1}{n}$
- Correct answerB$\dfrac{1}{n + 1}$
- C$\dfrac{1}{n - 1}$
- D$\dfrac{1}{n - 2}$
Solution
$I_{n+2} + I_n = \int\limits_0^{\pi/4} \tan^{n+2} x\ dx + \int\limits_0^{\pi/4} \tan^n x\ dx$
$I_{n+2} + I_n = \int\limits_0^{\pi/4} \tan^n x (\tan^2 x + 1)\ dx$
$I_{n+2} + I_n = \int\limits_0^{\pi/4} \tan^n x \sec^2 x\ dx$
Substituting $t = \tan x$, we get $dt = \sec^2 x\ dx$.
When $x = 0$, $t = 0$ and when $x = \pi/4$, $t = 1$.
$I_{n+2} + I_n = \int\limits_0^1 t^n\ dt$
$I_{n+2} + I_n = \left[ \dfrac{t^{n+1}}{n+1} \right]_0^1$
$I_{n+2} + I_n = \dfrac{1}{n+1}$
Answer: $\dfrac{1}{n + 1}$
$I_{n+2} + I_n = \int\limits_0^{\pi/4} \tan^n x (\tan^2 x + 1)\ dx$
$I_{n+2} + I_n = \int\limits_0^{\pi/4} \tan^n x \sec^2 x\ dx$
Substituting $t = \tan x$, we get $dt = \sec^2 x\ dx$.
When $x = 0$, $t = 0$ and when $x = \pi/4$, $t = 1$.
$I_{n+2} + I_n = \int\limits_0^1 t^n\ dt$
$I_{n+2} + I_n = \left[ \dfrac{t^{n+1}}{n+1} \right]_0^1$
$I_{n+2} + I_n = \dfrac{1}{n+1}$
Answer: $\dfrac{1}{n + 1}$
Hence, the correct answer is option: 'B'.