If-I-n-0-pi-4-tan-n-x-dx-n-N-then-I-n-2-I-n-

Question Number 69734 by mhmd last updated on 27/Sep/19

If I_{n} = \underset{0}{\int^{π / 4}} \tan^{n} x d x, n \in N, then I_{n + 2} + I_{n} =

Commented by mathmax by abdo last updated on 16/Oct/19

I_{n + 2} + I_{n} = \int_{0}^{\frac{π}{4}} ({t a n}^{n + 2} x + {t a n}^{n} x) d x = \int_{0}^{\frac{π}{4}} {t a n}^{n} x (1 + {t a n}^{2} x) d x

= {[\frac{1}{n + 1} {t a n}^{n + 1} x]}_{0}^{\frac{π}{4}} = \frac{1}{n + 1}