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

Answered by aleks041103 last updated on 09/Jun/22

$$\sqrt[1/4]{{tg}^{n^2-n+\frac{1}{4}}\left(x\right)}={tg}^{4n^2-4n+1}\left(x\right)=$$$$={tg}^{{(2n-1)}^2}\left(x\right)$$$$sincetg\left(x\right)hasaperiodof\pi$$$$\Rightarrow I=n{\int }_0^{\pi }{tg}^{{(2n-1)}^2}\left(x\right)dx=n{\int }_{-\pi /2}^{\pi /2}{tg}^{{(2n-1)}^2}\left(x\right)dx$$$$buttg\left(x\right)isanoddfunction.$$$$since{(2n-1)}^{2}isodd,then$$$${tg}^{{(2n-1)}^2}\left(x\right)isoddtoo.$$$$\Rightarrow I=n{\int }_{-a}^{a}odd\left(x\right)dx=0$$$$\Rightarrow I=0$$

Commented by Mastermind last updated on 09/Jun/22

$$Thanks$$

Commented by Mastermind last updated on 09/Jun/22

$$Thanks,but{what}^{\prime }sthemeaningoftg?$$

Commented by aleks041103 last updated on 09/Jun/22

$$tg\equiv tan$$

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