0-pi-2-tanx-1-7-dx-

Question Number 122176 by Dwaipayan Shikari last updated on 14/Nov/20

\int_{0}^{\frac{π}{2}} \sqrt[7]{t a n x} d x

Commented by Dwaipayan Shikari last updated on 14/Nov/20

I h a v e f o u n d \frac{π}{2} c o s e c (\frac{3 π}{7})

Commented by mnjuly1970 last updated on 14/Nov/20

p e r f e c t m r d w a i p a y a n . .

Answered by mnjuly1970 last updated on 14/Nov/20

I = \frac{1}{2} * 2 \int_{0}^{\frac{π}{2}} {s i n}^{\frac{1}{7}} (x) {c o s}^{\frac{- 1}{7}} d x

= \frac{1}{2} β (\frac{4}{7}, \frac{3}{7}) = \frac{1}{2} * \frac{Γ (\frac{4}{7}) Γ (1 - \frac{4}{7})}{Γ (1)}

= \frac{1}{2} * \frac{π}{s i n (\frac{4 π}{7})} = \frac{π}{2} c s c (\frac{4 π}{7}) ✓

Commented by Dwaipayan Shikari last updated on 15/Nov/20

W i t h p l e a s u r e

Answered by mathmax by abdo last updated on 14/Nov/20

I = \int_{0}^{\frac{π}{2}} {(tanx)}^{\frac{1}{7}} dx \Rightarrow I = \int_{0}^{\frac{π}{2}} \sin^{\frac{1}{7}} x \cos^{- \frac{1}{7}} xdx \Rightarrow

2 I = 2 \int_{0}^{\frac{π}{2}} \cos^{- \frac{1}{7}} x \sin^{\frac{1}{7}} xdx we know that

2 \int_{0}^{\frac{π}{2}} \cos^{2 p - 1} x \sin^{2 q - 1} xdx = B (p, q) = \frac{Γ (p) Γ (q)}{Γ (p + q)}

2 p - 1 = - \frac{1}{7} \Rightarrow 2 p = 1 - \frac{1}{7} = \frac{6}{7} \Rightarrow p = \frac{3}{7}

2 q - 1 = \frac{1}{7} \Rightarrow 2 q = 1 + \frac{1}{7} = \frac{8}{7} \Rightarrow q = \frac{4}{7} \Rightarrow

2 I = \frac{Γ (\frac{3}{7}) . Γ (\frac{4}{7})}{Γ (1)} \Rightarrow I = \frac{1}{2} Γ (\frac{3}{7}) . Γ (1 - \frac{3}{7}) = \frac{1}{2} . \frac{π}{\sin (\frac{π}{7})} \Rightarrow

I = \frac{π}{2 \sin (\frac{π}{7})}

Commented by mathmax by abdo last updated on 14/Nov/20

sorry I = \frac{π}{2 \sin (\frac{3 π}{7})}

Commented by Dwaipayan Shikari last updated on 15/Nov/20

T h a n k i n g y o u

Commented by mathmax by abdo last updated on 15/Nov/20

you are welcome