Q64 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
$\text { If } I_n=\int_\limits0^{\dfrac{\pi}{4}} \tan ^n x d x \text {, for } n \geq 2 \text {, then } I_n+I_{n-2}=$
- A$\dfrac{1}{n+1}$
- B$\dfrac{1}{n}$
- C$\dfrac{1}{n}+\dfrac{1}{n-2}$
- Correct answerD$\dfrac{1}{n-1}$
Solution
Given $I_n = \int_{0}^{\dfrac{\pi}{4}} \tan^{n} x dx$.
Consider the sum $I_n + I_{n-2} = \int_{0}^{\dfrac{\pi}{4}} \tan^{n} x dx + \int_{0}^{\dfrac{\pi}{4}} \tan^{n-2} x dx$.
$I_n + I_{n-2} = \int_{0}^{\dfrac{\pi}{4}} \tan^{n-2} x (\tan^{2} x + 1) dx$.
Using the identity $\tan^{2} x + 1 = \sec^{2} x$, we have:
$I_n + I_{n-2} = \int_{0}^{\dfrac{\pi}{4}} \tan^{n-2} x \sec^{2} x dx$.
Let $u = \tan x$, then $du = \sec^{2} x dx$.
When $x = 0$, $u = 0$. When $x = \dfrac{\pi}{4}$, $u = 1$.
$I_n + I_{n-2} = \int_{0}^{1} u^{n-2} du$.
$I_n + I_{n-2} = \left[ \dfrac{u^{n-1}}{n-1} \right]_{0}^{1} = \dfrac{1}{n-1}$.
Answer: $\dfrac{1}{n-1}$
Consider the sum $I_n + I_{n-2} = \int_{0}^{\dfrac{\pi}{4}} \tan^{n} x dx + \int_{0}^{\dfrac{\pi}{4}} \tan^{n-2} x dx$.
$I_n + I_{n-2} = \int_{0}^{\dfrac{\pi}{4}} \tan^{n-2} x (\tan^{2} x + 1) dx$.
Using the identity $\tan^{2} x + 1 = \sec^{2} x$, we have:
$I_n + I_{n-2} = \int_{0}^{\dfrac{\pi}{4}} \tan^{n-2} x \sec^{2} x dx$.
Let $u = \tan x$, then $du = \sec^{2} x dx$.
When $x = 0$, $u = 0$. When $x = \dfrac{\pi}{4}$, $u = 1$.
$I_n + I_{n-2} = \int_{0}^{1} u^{n-2} du$.
$I_n + I_{n-2} = \left[ \dfrac{u^{n-1}}{n-1} \right]_{0}^{1} = \dfrac{1}{n-1}$.
Answer: $\dfrac{1}{n-1}$
Hence, the correct answer is option: 'D'.