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Question Number 46182 by Meritguide1234 last updated on 22/Oct/18

Commented by maxmathsup by imad last updated on 22/Oct/18
$we have 1−x^2 +x^4 −....=Σ_(n=0) ^∞ (−x^2 )^n =(1/(1+x^2 )) and cos^2 x +cos^4 x +cos^6 x +...=cos^2 x(1+cos^2 x +cos^4 x +....) =cos^2 x{(1/(1−cos^2 x))}=(1/(tan^2 x)) ⇒ I = ∫_0 ^(π/4) ((tan^2 x)/(1+x^2 ))dx by parts I =∫_0 ^(π/4) tan^2 x(Σ_(n=0) ^∞ (−1)^n x^n )ex =Σ_(n=0) ^∞ (−1)^n ∫_0 ^(π/4) x^n tan^2 x dt =Σ_(n=0) ^∞ (−1)^n A_n with A_n =∫_0 ^(π/4) x^n tan^2 x dx by parts A_n =[(1/(n+1)) x^(n+1) tan^2 x]_0 ^(π/4) −∫_0 ^(π/4) (x^(n+1) /(n+1)) (2tanx)(1+tan^2 x)dx =(1/(n+1))((π/4))^(n+1) −(2/(n+1)) ∫_0 ^(π/4) x^(n+1) tanx dx −(2/(n+1)) ∫_0 ^(π/4) x^(n+1) tan^3 t dt...be continued...$
$w e h a v e 1 - x^{2} + x^{4} - . . . . = \sum_{n = 0}^{\infty} {(- x^{2})}^{n} = \frac{1}{1 + x^{2}} a n d$ ${c o s}^{2} x + {c o s}^{4} x + {c o s}^{6} x + . . . = {c o s}^{2} x (1 + {c o s}^{2} x + {c o s}^{4} x + . . . .)$ $= {c o s}^{2} x {\frac{1}{1 - {c o s}^{2} x}} = \frac{1}{{t a n}^{2} x} \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{{t a n}^{2} x}{1 + x^{2}} d x b y p a r t s$ $I = \int_{0}^{\frac{π}{4}} {t a n}^{2} x (\sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}) e x = \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\frac{π}{4}} x^{n} {t a n}^{2} x d t$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} A_{n} w i t h A_{n} = \int_{0}^{\frac{π}{4}} x^{n} {t a n}^{2} x d x b y p a r t s$ $A_{n} = {[\frac{1}{n + 1} x^{n + 1} {t a n}^{2} x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} \frac{x^{n + 1}}{n + 1} (2 t a n x) (1 + {t a n}^{2} x) d x$ $= \frac{1}{n + 1} {(\frac{π}{4})}^{n + 1} - \frac{2}{n + 1} \int_{0}^{\frac{π}{4}} x^{n + 1} t a n x d x - \frac{2}{n + 1} \int_{0}^{\frac{π}{4}} x^{n + 1} {t a n}^{3} t d t . . . b e c o n t i n u e d . . .$
Commented by maxmathsup by imad last updated on 22/Oct/18

$w e h a v e 0 ⩽ x ⩽ \frac{π}{4} \Rightarrow 0 ⩽ {t a n}^{2} x ⩽ 1 \Rightarrow 0 ⩽ \frac{{t a n}^{2} x}{1 + x^{2}} ⩽ 1 \Rightarrow$ $0 ⩽ \int_{0}^{\frac{π}{4}} \frac{{t a n}^{2} x}{1 + x^{2}} d x ⩽ \int_{0}^{\frac{π}{4}} d x = \frac{π}{4} \Rightarrow 0 ⩽ I ⩽ \frac{π}{4} . . . . .$
Commented by Meritguide1234 last updated on 22/Oct/18

$v e r y g o o d$
Commented by maxmathsup by imad last updated on 22/Oct/18

$t h a n k y o u s i r .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18

	$N_{r} = \frac{1}{1 + x^{2}} D_{r} = \frac{{c o s}^{2} x}{1 - {c o s}^{2} x} = {c o t}^{2} x$ $\int_{0}^{\frac{π}{4}} \frac{1}{(1 + x^{2})} \times \frac{1}{{c o t}^{2} x} d x$ $\int_{0}^{\frac{π}{4}} \frac{{t a n}^{2} x}{1 + x^{2}} d x$
	Commented by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18

	Commented by Meritguide1234 last updated on 22/Oct/18

	$n e e d m o r e s p e c i f i c . . .$
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