Question Number 200934 by sonukgindia last updated on 26/Nov/23
Answered by Frix last updated on 27/Nov/23
$$\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\left(\mathrm{tan}\:\pi{x}\right)^{\frac{\mathrm{4}}{\mathrm{5}}} {dx}\:\overset{{t}=\left(\mathrm{cot}\:{x}\right)^{\frac{\mathrm{1}}{\mathrm{5}}} } {=} \\ $$$$=−\frac{\mathrm{5}}{\pi}\underset{\infty} {\overset{\mathrm{0}} {\int}}\:\frac{{dt}}{{t}^{\mathrm{10}} +\mathrm{1}}\:=\frac{\mathrm{5}}{\pi}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{10}} +\mathrm{1}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$