0-pi-2-tan-m-1-k-d-for-m-gt-1-

Question Number 91735 by frc2crc last updated on 02/May/20

\int_{0}^{π / 2} \sqrt[k]{\tan^{m} α} d α f o r m > 1

Commented by mathmax by abdo last updated on 03/May/20

I = \int_{0}^{\frac{π}{2}} {({t a n}^{m} α)}^{\frac{1}{k}} d α c h s n g e m e n t t a n α = x g i v e

I = \int_{0}^{\infty} {(x^{m})}^{\frac{1}{k}} \frac{d x}{1 + x^{2}} = \int_{0}^{\infty} \frac{x^{\frac{m}{k}}}{1 + x^{2}} d x c h a n g e m e n t x = u^{\frac{1}{2}} g i v e

I = \int_{0}^{\infty} \frac{u^{\frac{m}{2 k}}}{1 + u} \frac{1}{2} u^{\frac{1}{2} - 1} d u = \frac{1}{2} \int_{0}^{\infty} \frac{u^{\frac{m}{2 k} + \frac{1}{2} - 1}}{1 + u} d u

w e h a v e p r o v e d t h a t \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} d t = \frac{π}{s i n (π a)} i f o < a < 1 \Rightarrow

I = \frac{1}{2} \times \frac{π}{s i n (π (\frac{m}{2 k} + \frac{1}{2}))} = \frac{π}{2 s i n (\frac{m π}{2 k} + \frac{π}{2})} \Rightarrow

I = \frac{π}{2 c o s (\frac{m π}{2 k})}

\frac{m}{2 k} + \frac{1}{2} - 1 = \frac{m}{2 k} - \frac{1}{2} = \frac{1}{2} (\frac{m}{k} - 1) < 0 \Rightarrow m < k s o t h e c o n d i t i o n i s

1 < m < k