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Question: If \(I_{n} = \int_{0}^{\pi/4}{\tan^{n}xdx,n \in N,}\) then \(I_{n + 2} + I_{n}equals\)...

If In=∫0π/4tan⁡nxdx,n∈N,I_{n} = \int_{0}^{\pi/4}{\tan^{n}xdx,n \in N,} then In+2+InequalsI_{n + 2} + I_{n}equals

A

1n\frac{1}{n}

B

1n−1\frac{1}{n - 1}

C

1n+1\frac{1}{n + 1}

D

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

Answer

1n+1\frac{1}{n + 1}

Explanation

Solution

We have,

In+2+In=∫0π/4tan⁡n+2xdx+∫0π/4tan⁡nxdxI_{n + 2} + I_{n} = \int_{0}^{\pi/4}{}\tan^{n + 2}{}xdx + \int_{0}^{\pi/4}{\tan^{n}xdx}=∫0π/4tan⁡nx(1+tan⁡2x)dx=∫0π/4tan⁡nxsec⁡2xdx\int_{0}^{\pi/4}{\tan^{n}x\left( 1 + \tan^{2}x \right)dx = \int_{0}^{\pi/4}{\tan^{n}x\sec^{2}xdx}}= ∫01tndt,wheret=tan⁡x\int_{0}^{1}{t^{n}dt,wheret = \tan x}

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