find-the-general-formula-0-pi-2-tan-x-dx-

Question Number 97928 by M±th+et+s last updated on 10/Jun/20

f i n d t h e g e n e r a l f o r m u l a

\int_{0}^{\frac{π}{2}} {t a n}^{α} (x) d x

Answered by abdomathmax last updated on 10/Jun/20

I = \int_{0}^{\frac{π}{2}} \tan^{α} xdx changement tanx = t give

I = \int_{0}^{\infty} \frac{t^{α}}{1 + t^{2}} dt also changement t = u^{\frac{1}{2}}

give I = \int_{0}^{\infty} \frac{u^{\frac{α}{2}}}{1 + u} \frac{1}{2} u^{- \frac{1}{2}} du

= \int_{0}^{\infty} \frac{u^{\frac{α - 1}{2}}}{1 + u} du we hsve proved that \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t} dt

= \frac{π}{\sin (π a)} if 0 < a < 1 \Rightarrow

I = \int_{0}^{\infty} \frac{u^{\frac{α + 1}{2} - 1}}{1 + u} du = \frac{π}{\sin (\frac{π}{2} (α + 1))} = \frac{π}{\cos (\frac{π α}{2})}

so \int_{0}^{\frac{π}{2}} \tan^{α} xdx = \frac{π}{\cos (\frac{π α}{2})}

conditions! 0 < \frac{α + 1}{2} < 1 \Rightarrow

0 < α + 1 < 2 \Rightarrow - 1 < α < 1 to get the convergence

Commented by M±th+et+s last updated on 11/Jun/20

t h a n k y o u s i r

Commented by mathmax by abdo last updated on 11/Jun/20

you are welcome .