calculate A_n =∫_0 ^((nπ)/4) (dx/(3cos^4 x +3sin^4 x−1))


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Question Number 97839 by mathmax by abdo last updated on 10/Jun/20

$calculate A_{n} = \int_{0}^{\frac{n π}{4}} \frac{dx}{{3 \cos}^{4} x + {3 \sin}^{4} x - 1}$
Commented by mr W last updated on 10/Jun/20

$3 (\cos^{4} x + \sin^{4} x) - 1$ $= 3 (\cos^{4} x + \sin^{4} x + 2 \cos^{2} x \sin^{2} x) - 1 - \frac{3}{2} \sin^{2} 2 x$ $= 3 {(\cos^{2} x + \sin^{2} x)}^{2} - 1 - \frac{3}{2} \sin^{2} 2 x$ $= \frac{1}{2} (4 - 3 \sin^{2} 2 x) ⩽ 2, ⩾ \frac{1}{2}$ $\Rightarrow \frac{1}{2} ⩽ \frac{1}{{3 \cos}^{4} x + 3 \sin^{4} x - 1} ⩽ 2$ $\Rightarrow \int_{0}^{\frac{n π}{4}} \frac{d x}{{3 \cos}^{4} x + 3 \sin^{4} x - 1} > \frac{1}{2} \times \frac{n π}{4} = \frac{n π}{8}$ $\Rightarrow \int_{0}^{\frac{n π}{4}} \frac{d x}{{3 \cos}^{4} x + 3 \sin^{4} x - 1} < 2 \times \frac{n π}{4} = \frac{n π}{2}$ $\Rightarrow \frac{n π}{8} < A_{n} < \frac{n π}{2}$ $i . e . A_{n} i s a l w a y s > 0, c a n n e v e r b e z e r o!$ $A_{n} = \int_{0}^{\frac{n π}{4}} \frac{2 d x}{4 - 3 \sin^{2} 2 x}$ $= \int_{0}^{\frac{n π}{2}} \frac{d t}{4 - 3 \sin^{2} t} w i t h t = 2 x$ $= n \int_{0}^{\frac{π}{2}} \frac{d t}{4 - 3 \sin^{2} t}$ $= n {[\frac{1}{2} \tan^{- 1} \frac{\tan t}{2}]}_{0}^{\frac{π}{2}}$ $= \frac{n}{2} \times \frac{π}{2}$ $= \frac{n π}{4}$
	Answered by smridha last updated on 10/Jun/20

	$f o r n o d d A_{n} = \frac{π}{4}$ $a n d f o r e v e n n A_{n} = 0$
	Answered by smridha last updated on 10/Jun/20

	$\int_{0}^{\frac{n π}{4}} \frac{d (t a n (2 x))}{2^{2} + {(t a n (2 x))}^{2}} = \frac{1}{2} {(\tan^{- 1} [\frac{t a n (2 x)}{2}])}_{0}^{\frac{n π}{4}}$ $= \frac{1}{2} (\tan^{- 1} [\frac{t a n (\frac{n π}{2})}{2}])$ $n o w f o r n = 2 m (e v e n) A_{n} = 0$ $n = 2 m + 1 (o d d) A_{n} = \frac{π}{4}$
	Answered by abdomathmax last updated on 10/Jun/20
	$we have 3(cos^4 x+sin^4 x)−1 =3{ (cos^2 x +sin^2 x)^2 −2cos^2 x sin^2 x}−1 =3{1−(1/2)sin^2 (2x)}−1 =2−(3/2)×((1−cos(4x))/2) =2−(3/4) +(3/4)cos(4x) =(5/4) +(3/4) cos(4x) ⇒ A_n =4∫_0 ^((nπ)/4) (dx/(5 +3cos(4x))) =_(4x=t) 4 ∫_0 ^(nπ) (dt/(4(5+3cost))) =∫_0 ^(nπ) (dt/(5+3cost)) =_(tan((t/2))=u) ∫_0 ^(tan(((nπ)/2))) ((2du)/((1+u^2 )(5+3((1−u^2 )/(1+u^2 ))))) =2∫_0 ^(tan(((nπ)/2))) (du/(5+5u^2 +3−3u^2 )) =2∫_0 ^(tan(((nπ)/2))) (du/(8+2u^2 )) =∫_0 ^(tan(((nπ)/2))) (du/(4+u^2 )) =_(u=2z) ∫_0 ^((1/2)tan(((nπ)/2))) ((2dz)/(4(1+z^2 ))) =(1/2) [arctan(z)]_0 ^((1/2)tan(((nπ)/2))) =(1/2) arctan((1/2)tan(((nπ)/2))) ⇒ A_(2n) =0 and A_(2n+1) =(1/2)×(π/2) =(π/4)$
	$we have 3 (\cos^{4} x + \sin^{4} x) - 1$ $= 3 {{(\cos^{2} x + \sin^{2} x)}^{2} - {2 \cos}^{2} x \sin^{2} x} - 1$ $= 3 {1 - \frac{1}{2} \sin^{2} (2 x)} - 1 = 2 - \frac{3}{2} \times \frac{1 - \cos (4 x)}{2}$ $= 2 - \frac{3}{4} + \frac{3}{4} \cos (4 x) = \frac{5}{4} + \frac{3}{4} \cos (4 x) \Rightarrow$ $A_{n} = 4 \int_{0}^{\frac{n π}{4}} \frac{dx}{5 + 3 \cos (4 x)}$ $=_{4 x = t} 4 \int_{0}^{n π} \frac{dt}{4 (5 + 3 cost)} = \int_{0}^{n π} \frac{dt}{5 + 3 cost}$ $=_{\tan (\frac{t}{2}) = u} \int_{0}^{\tan (\frac{n π}{2})} \frac{2 du}{(1 + u^{2}) (5 + 3 \frac{1 - u^{2}}{1 + u^{2}})}$ $= 2 \int_{0}^{\tan (\frac{n π}{2})} \frac{du}{5 + {5 u}^{2} + 3 - {3 u}^{2}} = 2 \int_{0}^{\tan (\frac{n π}{2})} \frac{du}{8 + {2 u}^{2}}$ $= \int_{0}^{\tan (\frac{n π}{2})} \frac{du}{4 + u^{2}} =_{u = 2 z} \int_{0}^{\frac{1}{2} \tan (\frac{n π}{2})} \frac{2 dz}{4 (1 + z^{2})}$ $= \frac{1}{2} {[\arctan (z)]}_{0}^{\frac{1}{2} \tan (\frac{n π}{2})}$ $= \frac{1}{2} \arctan (\frac{1}{2} \tan (\frac{n π}{2})) \Rightarrow$ $A_{2 n} = 0 and$ $A_{2 n + 1} = \frac{1}{2} \times \frac{π}{2} = \frac{π}{4}$
	Commented by smridha last updated on 10/Jun/20

	$t h a n k s a l o t f o r c o n f o r m a t i o n .$
	Commented by mr W last updated on 10/Jun/20

	$p r o f . a b d o s i r :$ $t h e r e s u l t A_{2 n} = 0 a n d A_{2 n + 1} = \frac{π}{4} i s$ $d e f i n i t i v e l y n o t c o r r e c t . a s i s h o w e d$ $a b o v e, \frac{n π}{8} < A_{n} < \frac{n π}{2}, t h i s i s f o r s u r e .$ $i f f (x) ⩾ \frac{1}{2}, \int_{a}^{b} f (x) d x c a n n e v e r b e 0!$ $i t h i n k w e {d o n}^{'} t n e e d t o d i s c u s s t h i s .$ $t h e q u e s t i o n i s w h e r e i s t h e m i s t a k e$ $i n y o u r w o r k i n g . i t h i n k t h e m i s t a k e$ $l i e s i n t h e u s e o f f u n c t i o n s \tan x a n d$ $\tan^{- 1} x . w e m u s t b e v e r y c a r e f u l w i t h$ $t h e m, b e c a u s e \tan x i s n o t c o n t i n u o u s .$ $i l e t y o u t h i n k a b o u t t h a t . y o u k n o w$ $w h a t i m e a n .$ $t h e c o r r e c t w a y t o t r e a t t h i s p r o b l e m$ $i s t o d i v i d e t h e i n t e r v a l o f i n t e g r a l$ $0 \to \frac{n π}{4} i n t o m a n y s m a l l s e g m e n t s,$ $w i t h i n e a c h o f t h e s e s e g m e n t s t h e$ $f u n c t i o n i s c o n t i n u o u s . i . e . w e t a k e$ $0 \to \frac{π}{4}$ $\frac{π}{4} \to \frac{2 π}{4}$ $\frac{2 π}{4} \to \frac{3 π}{4}$ $. . .$ $\frac{(n - 1) π}{4} \to \frac{n π}{4}$ $y o u w i l l s e e i n e a c h o f t h e s e s e g m e n t s$ $t h e i n t e g r a l d e l i v e r s t h e r e s u l t \frac{π}{4},$ $t h e r e f o r e t h e t o t a l r e s u l t i s n \times \frac{π}{4} .$ $A_{n} = \int_{0}^{\frac{n π}{4}} f (x) d x$ $= \int_{0}^{\frac{π}{4}} f (x) d x + \int_{\frac{π}{4}}^{\frac{2 π}{4}} f (x) d x + . . . + \int_{\frac{(n - 1) π}{4}}^{\frac{n π}{4}} f (x) d x$ $= \frac{π}{4} + \frac{π}{4} + . . . + \frac{π}{4}$ $= n \times \frac{π}{4}$ $= \frac{n π}{4}$
	Commented by mr W last updated on 10/Jun/20

	Commented by mathmax by abdo last updated on 10/Jun/20

	$thank sir for this clarification$
	Commented by mr W last updated on 10/Jun/20

	$t h a n k s f o r t h e n i c e q u e s t i o n!$
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