calculate : Ω = ∫_0 ^( (π/4)) (( 1+ tan^( 4) (x))/(cot^( 2) (x))) dx=?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
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
	$Ω=∫_0 ^(π/4) ((1+tan^4 (x))/(cotan^2 (x)))dx = ∫_0 ^(π/4) {tan^2 x+tan^6 x}dx u=tanx→du=(1+tan^2 x)dx → dx=(du/(1+u^2 )) ⇒ Ω=∫_0 ^1 ((u^2 +u^6 )/(1+u^2 ))du=∫_0 ^1 ((u^2 +1−1)/(1+u^2 ))du+∫_0 ^1 (u^6 /(1+u^2 ))du Ω=∫_0 ^1 {1−(1/(1+u^2 ))}dx+∫_0 ^1 {u^4 −u^2 +1−(1/(1+u^2 ))}du ⇒ Ω = 1−(π/4)+(1/5)−(1/3)+1−(π/4) = ((28)/(15))−(π/2). Ω = ∫_0 ^1 ((1+tan^2 (x))/(cotan^2 (x)))dx = ((28)/(15))−(π/2) ................Le puissant...............$
	$Ω = \int_{0}^{\frac{π}{4}} \frac{1 + {t a n}^{4} (x)}{{c o t a n}^{2} (x)} d x = \int_{0}^{\frac{π}{4}} {{t a n}^{2} x + {t a n}^{6} x} d x$ $u = t a n x \to d u = (1 + {t a n}^{2} x) d x \to d x = \frac{d u}{1 + u^{2}}$ $\Rightarrow Ω = \int_{0}^{1} \frac{u^{2} + u^{6}}{1 + u^{2}} d u = \int_{0}^{1} \frac{u^{2} + 1 - 1}{1 + u^{2}} d u + \int_{0}^{1} \frac{u^{6}}{1 + u^{2}} d u$ $Ω = \int_{0}^{1} {1 - \frac{1}{1 + u^{2}}} d x + \int_{0}^{1} {u^{4} - u^{2} + 1 - \frac{1}{1 + u^{2}}} d u$ $\Rightarrow Ω = 1 - \frac{π}{4} + \frac{1}{5} - \frac{1}{3} + 1 - \frac{π}{4} = \frac{28}{15} - \frac{π}{2} .$ $Ω = \int_{0}^{1} \frac{1 + {t a n}^{2} (x)}{{c o t a n}^{2} (x)} d x = \frac{28}{15} - \frac{π}{2}$ $. . . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . . . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum