If I_{n} = \int ↙\stackrel{\pi \/ 4}{0}⁡\tan ^{n}\theta d\theta then the following reduction formula exists:

Correct Answer is : In+In−2=1n−1

I_{n} + I_{n−2} = \frac{1}{n−1}

I_{n} + I_{n−2} + \frac{1}{n + 1} = 0

I_{n}−I_{n−2} = \frac{1}{n + 1}

I_{n}−I_{n−2} + \frac{1}{n−1} = 0

AAI ATC Junior Executive 2016 Shift 2 Paper

Question : 117 of 120

Marks: +1, -0

I_{n} = ∫ ↙{0}↖{π \/ 4}⁡\tan ^{n}θ dθ

then the following reduction formula exists:

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