0-pi-2-tan-x-1-n-dx-

Question Number 133016 by metamorfose last updated on 18/Feb/21

\int_{0}^{\frac{π}{2}} {(t a n (x))}^{\frac{1}{n}} d x \dots

Answered by Ar Brandon last updated on 18/Feb/21

I = \int_{0}^{\frac{π}{2}} {(tanx)}^{\frac{1}{n}} dx = \int_{0}^{\frac{π}{2}} {(sinx)}^{\frac{1}{n}} {(cosx)}^{- \frac{1}{n}} dx

= \frac{1}{2} Γ (\frac{n + 1}{2 n}) Γ (\frac{n - 1}{2 n}) = \frac{1}{2} \cdot \frac{π}{\sin (\frac{n + 1}{2 n} π)}, n > 1

Answered by Dwaipayan Shikari last updated on 18/Feb/21

\int_{0}^{\frac{π}{2}} \sqrt[n]{t a n x} d x = \frac{Γ (\frac{1}{2} - \frac{1}{2 n}) Γ (\frac{1}{2} + \frac{1}{2 n})}{2} = \frac{π}{2 s i n (\frac{π}{2} + \frac{π}{2 n})}

Answered by mathmax by abdo last updated on 20/Feb/21

I = \int_{0}^{\frac{π}{2}} {(tanx)}^{\frac{1}{n}} dx changement tanx = t give

I = \int_{0}^{\infty} \frac{t^{\frac{1}{n}}}{1 + t^{2}} dt =_{t = z^{\frac{1}{2}}} \int_{0}^{\infty} \frac{z^{\frac{1}{2 n}}}{1 + z} \frac{1}{2} z^{- \frac{1}{2}} dz = \frac{1}{2} \int_{0}^{\infty} \frac{z^{\frac{1}{2 n} + \frac{1}{2} - 1}}{1 + z} dz

= \frac{1}{2} \frac{π}{\sin (π (\frac{1}{2 n} + \frac{1}{2}))} = \frac{π}{2 \sin (\frac{π}{2 n} + \frac{π}{2})} = \frac{π}{2 \cos (\frac{π}{2 n})} (with n > 1)