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If $u_{n}\int_{0}^{\pi/4}\tan^{n}\theta d\theta$ , then u_n + u_n-2 is

Answer:

Explanation:

Given :  $u_{n}\int_{0}^{\pi/4}\tan^{n}\theta d\theta$

= $\int_{0}^{\pi/4}\tan^{2}\theta \tan^{n-2}\theta d\theta$

= $\int_{0}^{\pi/4}(\sec^{2}\theta-1) \tan^{n-2}\theta d\theta$

= $\int_{0}^{\pi/4}\sec^{2}\theta \tan^{n-2}\theta d\theta$ - $\int_{0}^{\pi/4}\tan^{n-2}\theta d\theta$

= $\int_{0}^{\pi/4}(\sec^{2}\theta) \tan^{n-2}\theta d\theta$ - u_n-2

u_n + u_n-2 = $\int_{0}^{\pi/4}\sec^{2}\theta\tan^{n-2}\theta d\theta$

= $\frac{\tan^{n-1}\theta}{n-1}\mid_0^\frac{\pi}{4}$ = $\frac{1}{n-1}$