1. Tardigrade
  2. Question
  3. Mathematics
  4. If un=∫ limits0π / 4 tan n θ d θ then un+un-2 is:

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. If $u_{n}=\int\limits_{0}^{\pi / 4} \tan ^{n} \theta d \theta$ then $u_{n}+u_{n-2}$ is:

VITEEEVITEEE 2017

Solution:

Given $: u_{n}=\int\limits_{0}^{\pi / 4} \tan ^{n} \theta d \theta$
$=\int\limits_{0}^{\pi / 4} \tan ^{2} \theta \tan ^{n-2} \theta d \theta$
$=\int\limits_{0}^{\pi / 4}\left(\sec ^{2} \theta-1\right) \tan ^{n-2} \theta d \theta$
$=\int\limits_{0}^{\pi / 4} \sec ^{2} \theta \tan ^{n-2} \theta d \theta$
$-\int\limits_{0}^{\pi / 4} \tan ^{n-2} \theta d \theta$
$=\int\limits_{0}^{\pi / 4} \sec ^{2} \theta \tan ^{n-2} \theta d \theta-u_{n-2}$
$\Rightarrow u_{n}+u_{n-2}=\int\limits_{0}^{\pi / 4} \sec ^{2} \theta t \tan ^{n-2} \theta d \theta$
$=\left.\frac{\tan ^{n-1} \theta}{n-1}\right|_{0} ^{\pi / 4}=\frac{1}{n-1}$