0-pi-2-dx-1-tan-4-x-

Question Number 114699 by bemath last updated on 20/Sep/20

\underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\sqrt{1 + \tan^{4} x}} ?

Commented by Dwaipayan Shikari last updated on 20/Sep/20

Answered by bobhans last updated on 20/Sep/20

r e p l a c i n g x = \frac{π}{2} - x

I = \underset{\frac{π}{2}}{\int^{0}} \frac{- d x}{\sqrt{1 + \tan^{4} (\frac{π}{2} - x)}}

I = \underset{0}{\int^{\frac{π}{2}}} \frac{d x}{\sqrt{1 + \cot^{4} x}} = \underset{0}{\int^{\frac{π}{2}}} \frac{\tan^{2} x}{\sqrt{\tan^{4} x + 1}} d x

2 I = \underset{0}{\int^{\frac{π}{2}}} \frac{\sec^{2} x}{\sqrt{\tan^{4} x + 1}} d x

I = \frac{1}{2} {\int^{\infty}}_{1} \frac{d t}{\sqrt{t^{4} + 1}}; [t = \tan x]

s e t q = 1 + t^{4}; t = {(q - 1)}^{\frac{1}{4}}

I = \frac{1}{2} {\int^{\infty}}_{1} \frac{1}{\sqrt{q}} . \frac{1}{4} {(q - 1)}^{- \frac{3}{4}} d q

I = \frac{1}{8} \underset{1}{\int^{\infty}} q^{- \frac{1}{2}} {(q - 1)}^{- \frac{3}{4}} d q

I = \frac{1}{8} {\int^{\infty}}_{1} q^{- \frac{5}{4}} {(1 - q^{- 1})}^{- \frac{3}{4}} d q

I = \frac{1}{8} . \frac{Γ^{2} (\frac{1}{4})}{Γ (\frac{1}{2})} = \frac{1}{8 \sqrt{π}} . Γ^{2} (\frac{1}{4})

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