∫_( 0) ^(π/2) ((x+sin x)/(1+cos x)) dx =


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Question Number 53693 by gunawan last updated on 25/Jan/19

$\underset{0}{\int^{π / 2}} \frac{x + \sin x}{1 + \cos x} d x =$
Commented by maxmathsup by imad last updated on 25/Jan/19

$l e t I = \int_{0}^{\frac{π}{2}} \frac{x + s i n x}{1 + c o s x} \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{x}{1 + c o s x} d x + \int_{0}^{\frac{π}{2}} \frac{s i n x}{1 + c o s x} d x b u t$ $\int_{0}^{\frac{π}{2}} \frac{s i n x}{1 + c o s x} d x = - \int_{0}^{\frac{π}{2}} \frac{{(c o s x)}^{,}}{1 + c o s x} d x = - {[l n ∣ 1 + c o s x ∣]}_{0}^{\frac{π}{2}} = - (- l n (2)) = l n (2)$ $\int_{0}^{\frac{π}{2}} \frac{x}{1 + c o s x} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{x}{{c o s}^{2} (\frac{x}{2})} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} x (1 + {t a n}^{2} (\frac{x}{2}) d x b u t$ $\int_{0}^{\frac{π}{2}} x (1 + {t a n}^{2} (\frac{x}{2})) d x =_{\frac{x}{2} = t} \int_{0}^{\frac{π}{4}} (2 t) (1 + {t a n}^{2} t) (2) d t$ $= 4 \int_{0}^{\frac{π}{4}} t (1 + {t a n}^{2} t) d t b y p a r t s u = t a n d v^{^{'}} = 1 + {t a n}^{2} t \Rightarrow$ $\int_{0}^{\frac{π}{4}} t (1 + {t a n}^{2} t) d t = {[t t a n t]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} t a n t d t$ $= \frac{π}{4} + \int_{0}^{\frac{π}{4}} \frac{- s i n t}{c o s t} d t = \frac{π}{4} + {[l n ∣ c o s t ∣]}_{0}^{\frac{π}{4}} = \frac{π}{4} + l n (\frac{1}{\sqrt{2}}) = \frac{π}{4} - \frac{1}{2} l n (2) \Rightarrow$ $\int_{0}^{\frac{π}{2}} x (1 + {t a n}^{2} (\frac{x}{2})) d x = π - 2 l n (2) \Rightarrow$ $I = l n (2) + \frac{1}{2} (π - 2 l n (2)) \Rightarrow I = \frac{π}{2} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19
	$∫_0 ^(π/2) (x/(1+cosx))dx−∫_0 ^(π/2) ((d(1+cosx))/(1+cosx)) ∫_0 ^(π/2) x×(1/(2cos^2 (x/2)))−∫_0 ^(π/2) ((d(1+cosx))/(1+cosx)) now... ∫xsec^2 (x/2)dx x×((tan(x/2))/(1/2))−∫[(dx/dx)∫sec^2 (x/2)dx]dx =2xtan(x/2)−∫((tan(x/2))/(1/2))dx =2xtan(x/2)−2lnsec(x/2) so (1/2)∫_0 ^(π/2) xsec^2 (x/2)−∫_0 ^(π/2) ((d(1+cosx))/(1+cosx)) (1/2)∣2xtan(x/2)−2lnsec(x/2)∣_0 ^(π/2) −∣ln(1+cosx)∣_0 ^(π/2) ={((π/2)tan(π/4)−lnsec(π/4))−(0×tan(0/2)−lnsec0)}−{ln(1+0)−ln(1+1)} ={((π/2)×1−ln(√2) )−(0−0)}−{0−ln2} =(π/2)−(1/2)ln2+ln2 =(π/2)+(1/2)ln2 pls check steps...$
	$\int_{0}^{\frac{π}{2}} \frac{x}{1 + c o s x} d x - \int_{0}^{\frac{π}{2}} \frac{d (1 + c o s x)}{1 + c o s x}$ $\int_{0}^{\frac{π}{2}} x \times \frac{1}{2 {c o s}^{2} \frac{x}{2}} - \int_{0}^{\frac{π}{2}} \frac{d (1 + c o s x)}{1 + c o s x}$ $n o w . . .$ $\int {x s e c}^{2} \frac{x}{2} d x$ $x \times \frac{t a n \frac{x}{2}}{\frac{1}{2}} - \int [\frac{d x}{d x} \int {s e c}^{2} \frac{x}{2} d x] d x$ $= 2 x t a n \frac{x}{2} - \int \frac{t a n \frac{x}{2}}{\frac{1}{2}} d x$ $= 2 x t a n \frac{x}{2} - 2 l n s e c \frac{x}{2}$ $s o$ $\frac{1}{2} \int_{0}^{\frac{π}{2}} {x s e c}^{2} \frac{x}{2} - \int_{0}^{\frac{π}{2}} \frac{d (1 + c o s x)}{1 + c o s x}$ $\frac{1}{2} ∣ 2 x t a n \frac{x}{2} - 2 l n s e c \frac{x}{2} ∣_{0}^{\frac{π}{2}} - ∣ l n (1 + c o s x) ∣_{0}^{\frac{π}{2}}$ $= {(\frac{π}{2} t a n \frac{π}{4} - l n s e c \frac{π}{4}) - (0 \times t a n \frac{0}{2} - l n s e c 0)} - {l n (1 + 0) - l n (1 + 1)}$ $= {(\frac{π}{2} \times 1 - l n \sqrt{2}) - (0 - 0)} - {0 - l n 2}$ $= \frac{π}{2} - \frac{1}{2} l n 2 + l n 2$ $= \frac{π}{2} + \frac{1}{2} l n 2$ $p l s c h e c k s t e p s . . .$
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