please help integrate ((x+sin x)/(1+cos x))dx


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Question Number 50633 by Necxx last updated on 18/Dec/18

$p l e a s e h e l p i n t e g r a t e \frac{x + \sin x}{1 + \cos x} d x$
Commented by maxmathsup by imad last updated on 18/Dec/18
$let I = ∫ ((x+sinx)/(1+cosx))dx we have I =∫ (x/(1+cosx))dx +∫ ((sinx)/(1+cosx))dx but ∫ ((sinx)/(1+cosx))dx =∫ ((−d(cosx))/(1+cosx)) dx=−ln∣1+cosx∣ +c_1 changement tan((x/2))=t give ∫ (x/(1+cosx))dx = ∫ ((2arctant)/(1+((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =4 ∫ ((arctan(t))/(1+t^2 +1−t^2 ))dt =2 ∫ arctan(t)dt =_(by parts) 2{ t arctant−∫ (t/(1+t^2 )) dt} =2t arctan(t) −∫ ((2t)/(1+t^2 ))dt =2t arctan(t) −ln(1+t^2 ) +c_2 =2tan((x/2))(x/2) −ln(1+tan^2 ((x/2))+c_2 =x tan((x/2))−ln(1+tan^2 ((x/2))) +c_2 finally I =−ln∣1+cosx∣ +xtan((x/2))−ln(1+tan^2 ((x/2))) +C$
$l e t I = \int \frac{x + s i n x}{1 + c o s x} d x w e h a v e I = \int \frac{x}{1 + c o s x} d x + \int \frac{s i n x}{1 + c o s x} d x b u t$ $\int \frac{s i n x}{1 + c o s x} d x = \int \frac{- d (c o s x)}{1 + c o s x} d x = - l n ∣ 1 + c o s x ∣ + c_{1}$ $c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $\int \frac{x}{1 + c o s x} d x = \int \frac{2 a r c t a n t}{1 + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 4 \int \frac{a r c t a n (t)}{1 + t^{2} + 1 - t^{2}} d t$ $= 2 \int a r c t a n (t) d t =_{b y p a r t s} 2 {t a r c t a n t - \int \frac{t}{1 + t^{2}} d t}$ $= 2 t a r c t a n (t) - \int \frac{2 t}{1 + t^{2}} d t = 2 t a r c t a n (t) - l n (1 + t^{2}) + c_{2}$ $= 2 t a n (\frac{x}{2}) \frac{x}{2} - l n (1 + {t a n}^{2} (\frac{x}{2}) + c_{2}$ $= x t a n (\frac{x}{2}) - l n (1 + {t a n}^{2} (\frac{x}{2})) + c_{2} f i n a l l y$ $I = - l n ∣ 1 + c o s x ∣ + x t a n (\frac{x}{2}) - l n (1 + {t a n}^{2} (\frac{x}{2})) + C$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18

	$\int \frac{x + 2 s i n (\frac{x}{2}) c o s (\frac{x}{2})}{2 {c o s}^{2} (\frac{x}{2})} d x$ $\int [\frac{{x s e c}^{2} (\frac{x}{2})}{2} + t a n (\frac{x}{2})] d x$ $\int [x . \frac{d}{d x} (t a n \frac{x}{2}) + t a n (\frac{x}{2}) \frac{d x}{d x}] d x$ $\int \frac{d}{d x} (x t a n \frac{x}{2}) d x$ $\int d (x t a n \frac{x}{2})$ $x t a n \frac{x}{2} + c$
	Commented by ajfour last updated on 18/Dec/18

	$V e r y N i c e S i r!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 18/Dec/18

	$t h a n k y o u s i r . . .$
	Commented by Necxx last updated on 18/Dec/18

	$t h a n k y o u s o m u c h s i r$
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