Integration Questions

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

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

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

Answered by bobhans last updated on 20/Sep/20

$${replacing}\:{x}\:=\:\frac{\pi}{\mathrm{2}}−{x} \\$$$${I}=\underset{\frac{\pi}{\mathrm{2}}} {\overset{\mathrm{0}} {\int}}\:\frac{−{dx}}{\:\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}} \\$$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{dx}}{\:\sqrt{\mathrm{1}+\mathrm{cot}\:^{\mathrm{4}} {x}}}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}}{\:\sqrt{\mathrm{tan}\:^{\mathrm{4}} {x}+\mathrm{1}}}{dx} \\$$$$\mathrm{2}{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{\mathrm{sec}\:^{\mathrm{2}} {x}}{\:\sqrt{\mathrm{tan}\:^{\mathrm{4}} {x}+\mathrm{1}}}\:{dx}\: \\$$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\int}_{\mathrm{1}} \:\frac{{dt}}{\:\sqrt{{t}^{\mathrm{4}} +\mathrm{1}}}\:;\:\left[\:{t}\:=\:\mathrm{tan}\:{x}\:\right] \\$$$${set}\:{q}\:=\:\mathrm{1}+{t}^{\mathrm{4}} \:;\:{t}\:=\:\left({q}−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\$$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\overset{\infty} {\int}_{\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{q}}}.\frac{\mathrm{1}}{\mathrm{4}}\left({q}−\mathrm{1}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} \:{dq} \\$$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\underset{\mathrm{1}} {\overset{\infty} {\int}}\:{q}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left({q}−\mathrm{1}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} \:{dq} \\$$$${I}=\:\frac{\mathrm{1}}{\:\mathrm{8}}\overset{\infty} {\int}_{\mathrm{1}} {q}^{−\frac{\mathrm{5}}{\mathrm{4}}} \left(\mathrm{1}−{q}^{−\mathrm{1}} \right)^{−\frac{\mathrm{3}}{\mathrm{4}}} {dq} \\$$$${I}=\frac{\mathrm{1}}{\mathrm{8}}.\frac{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{8}\sqrt{\pi}}.\Gamma^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right) \\$$