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

$$\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{dx}{\sqrt{1+\tan {}^{4}x}}?$$

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

Answered by bobhans last updated on 20/Sep/20

$$replacingx=\frac{\pi }{2}-x$$$$I=\underset{\frac{\pi }{2}}{\stackrel{0}{\int }}\frac{-dx}{\sqrt{1+\tan {}^4(\frac{\pi }{2}-x)}}$$$$I=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{dx}{\sqrt{1+\cot {}^{4}x}}=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\tan {}^{2}x}{\sqrt{\tan {}^{4}x+1}}dx$$$$2I=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{\sec {}^{2}x}{\sqrt{\tan {}^{4}x+1}}dx$$$$I=\frac{1}{2}{\stackrel{\mathrm{\infty }}{\int }}_1\frac{dt}{\sqrt{t^4+1}};[t=\tan x]$$$$setq=1+t^4;t={(q-1)}^{\frac{1}{4}}$$$$I=\frac{1}{2}{\stackrel{\mathrm{\infty }}{\int }}_1\frac{1}{\sqrt{q}}.\frac{1}{4}{(q-1)}^{-\frac{3}{4}}dq$$$$I=\frac{1}{8}\underset{1}{\stackrel{\mathrm{\infty }}{\int }}{q}^{-\frac{1}{2}}{(q-1)}^{-\frac{3}{4}}dq$$$$I=\frac{1}{8}{\stackrel{\mathrm{\infty }}{\int }}_1{q}^{-\frac{5}{4}}{(1-q^{-1})}^{-\frac{3}{4}}dq$$$$I=\frac{1}{8}.\frac{{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)}=\frac{1}{8\sqrt{\pi }}.{\mathrm{\Gamma }}^2\left(\frac{1}{4}\right)$$

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