Question-171179

Question Number 171179 by Mastermind last updated on 09/Jun/22

Answered by aleks041103 last updated on 09/Jun/22

((tg^(n^2 −n+(1/4)) (x)))^(1/(1/4)) =tg^(4n^2 −4n+1) (x)= =tg^((2n−1)^2 ) (x) since tg(x) has a period of π ⇒I=n∫_0 ^π tg^((2n−1)^2 ) (x)dx=n∫_(−π/2) ^(π/2) tg^((2n−1)^2 ) (x)dx but tg(x) is an odd function. since (2n−1)^2 is odd, then tg^((2n−1)^2 ) (x) is odd too. ⇒I=n∫_(−a) ^a odd(x) dx=0 ⇒I=0

$$\sqrt[{\mathrm{1}/\mathrm{4}}]{{tg}^{{n}^{\mathrm{2}} −{n}+\frac{\mathrm{1}}{\mathrm{4}}} \left({x}\right)}={tg}^{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{4}{n}+\mathrm{1}} \left({x}\right)= \\ $$$$={tg}^{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \left({x}\right) \\ $$$${since}\:{tg}\left({x}\right)\:{has}\:{a}\:{period}\:{of}\:\pi \\ $$$$\Rightarrow{I}={n}\int_{\mathrm{0}} ^{\pi} {tg}^{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \left({x}\right){dx}={n}\int_{−\pi/\mathrm{2}} ^{\pi/\mathrm{2}} {tg}^{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \left({x}\right){dx} \\ $$$${but}\:{tg}\left({x}\right)\:{is}\:{an}\:{odd}\:{function}. \\ $$$${since}\:\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} \:{is}\:{odd},\:{then} \\ $$$${tg}^{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \left({x}\right)\:{is}\:{odd}\:{too}. \\ $$$$\Rightarrow{I}={n}\int_{−{a}} ^{{a}} {odd}\left({x}\right)\:{dx}=\mathrm{0} \\ $$$$\Rightarrow{I}=\mathrm{0} \\ $$

Commented by Mastermind last updated on 09/Jun/22