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Question: In = \(\int_{0}^{\pi/4}\tan^{n}\)x dx, then \(\lim_{n \rightarrow \infty}\) n [I<sub>n</sub> + I<sub...

In = ∫0π/4tan⁡n\int_{0}^{\pi/4}\tan^{n}x dx, then lim⁡n→∞\lim_{n \rightarrow \infty} n [In + In + 2] equals -

A

12\frac{1}{2}

B

1

C

D

zero

Answer

1

Explanation

Solution

In + In + 2 =∫0π/4tan⁡nx(1+tan⁡2x)dx\int_{0}^{\pi/4}{\tan^{n}x(1 + \tan^{2}x)dx}

= ∫0π/4tan⁡nxsec⁡2xdx\int_{0}^{\pi/4}{\tan^{n}x\sec^{2}xdx}

= ∫01tndt\int_{0}^{1}{t^{n}dt}, where t = tan x

\ In + In + 2 = 1n+1\frac{1}{n + 1}

Ž lim⁡n→∞\lim_{n \rightarrow \infty}n [In + In + 2]

= lim⁡n→∞\lim_{n \rightarrow \infty}n . 1n+1\frac{1}{n + 1} = lim⁡n→∞\lim _ { n \rightarrow \infty } nn+1\frac{n}{n + 1}

= = 1.

Hence (2) is the correct answer.