calculate-0-pi-4-1-tan-4-x-cot-2-x-dx-

Question Number 158596 by mnjuly1970 last updated on 06/Nov/21

c a l c u l a t e :

Ω = \int_{0}^{\frac{π}{4}} \frac{1 + {t a n}^{4} (x)}{{c o t}^{2} (x)} d x = ?

Answered by puissant last updated on 06/Nov/21

Ω = \int_{0}^{\frac{π}{4}} \frac{1 + {t a n}^{4} (x)}{{c o t a n}^{2} (x)} d x = \int_{0}^{\frac{π}{4}} {{t a n}^{2} x + {t a n}^{6} x} d x

u = t a n x \to d u = (1 + {t a n}^{2} x) d x \to d x = \frac{d u}{1 + u^{2}}

\Rightarrow Ω = \int_{0}^{1} \frac{u^{2} + u^{6}}{1 + u^{2}} d u = \int_{0}^{1} \frac{u^{2} + 1 - 1}{1 + u^{2}} d u + \int_{0}^{1} \frac{u^{6}}{1 + u^{2}} d u

Ω = \int_{0}^{1} {1 - \frac{1}{1 + u^{2}}} d x + \int_{0}^{1} {u^{4} - u^{2} + 1 - \frac{1}{1 + u^{2}}} d u

\Rightarrow Ω = 1 - \frac{π}{4} + \frac{1}{5} - \frac{1}{3} + 1 - \frac{π}{4} = \frac{28}{15} - \frac{π}{2} .

Ω = \int_{0}^{1} \frac{1 + {t a n}^{2} (x)}{{c o t a n}^{2} (x)} d x = \frac{28}{15} - \frac{π}{2}

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