∫((sinx)/(1+sinx+sin2x))dx


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Question Number 66589 by Tanmay chaudhury last updated on 17/Aug/19

$\int \frac{s i n x}{1 + s i n x + s i n 2 x} d x$
Commented by MJS last updated on 17/Aug/19

$Weierstrass [t = \tan \frac{x}{2} \to d x = \frac{2 d t}{t^{2} + 1}] leads to$ $4 \int \frac{t}{(t + 1) (t^{3} - 3 t^{2} + 5 t + 1)} d t$ $t^{3} - 3 t^{2} + 5 t + 1 =$ $= (t - α - β - 1) (t^{2} + (α + β - 2) t + (α^{2} + β^{2} - α β - α - β + 1))$ $α = \sqrt[3]{- 2 - \frac{2}{9} \sqrt{87}} β = \sqrt[3]{- 2 + \frac{2}{9} \sqrt{87}}$ $α + β + 1 = A$ $α + β - 2 = B$ $α^{2} + β^{2} - α β - α - β + 1 = C$ $4 \int \frac{t}{(t + 1) (t - A) (t^{2} + B t + C)} d t =$ $= - \frac{4}{(A + 1) (B - C - 1)} \int \frac{d t}{t + 1} +$ $+ \frac{4 A}{(A + 1) (A^{2} + A B + C)} \int \frac{d t}{t - A} +$ $+ \frac{4}{(A^{2} + A B + C) (B - C - 1)} \int \frac{(A + C) t + (A + B - 1) C}{t^{2} + B t + C} d t$ $now solve these . . .$ $I found no easier path$
Commented by MJS last updated on 17/Aug/19

$. . . corrected a mistake$
Commented by Tanmay chaudhury last updated on 17/Aug/19

$e x c e l l e n t s i r . . . i s h a l l t r y t o c o m p l e t e$
Commented by mathmax by abdo last updated on 18/Aug/19
$let I =∫ ((sinx)/(1+sinx +sin(2x)))dx ⇒ I =∫ ((sinx)/(1+sinx +2sinx cosx))dx changement tan((x/2))=t give I =∫ (((2t)/(1+t^2 ))/(1+((2t)/(1+t^2 ))+2 ((2t)/(1+t^2 ))((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫((4t)/((1+t^2 )^2 {1+((2t)/(1+t^2 )) +((4t(1−t^2 ))/((1+t^2 )^2 ))}))dt =∫ ((4tdt)/((1+t^2 )^2 +2t(1+t^2 )+4t(1−t^2 ))) =∫ ((4tdt)/(t^4 +2t^2 +1+2t+2t^3 +4t−4t^3 )) =∫ ((4tdt)/(t^4 −2t^3 +2t^2 +6t +1)) let decompose the fraction F(t) =((4t)/(t^4 −2t^3 +2t^2 +6t +1)) t^4 −2t^3 +2t^2 +6t +1 =0 the roots are t_1 =1,5898 +1,7445i (complex) t_2 ∼1,5898−1,7445i (complex) t_3 =−1 (real) t_4 ∼−0,1795 (real) ⇒ F(t)∼ ((4t)/((t−t_1 )(t−t_1 ^− )(t+1)(t−t_4 ))) =((4t)/((t^2 −2Re(t_1 )t +∣t_1 ∣^2 )(t+1)(t−t_4 ))) =(a/(t+1)) +(b/(t−t_4 )) +((ct+d)/(t^2 −2Re(t_1 )t +∣t_1 ∣^2 )) ⇒I =aln∣t+1∣+bln∣t−t_4 ∣ +∫ ((ct +d)/(t^2 −2Re(t_1 )t +∣t_1 ∣^2 )) rest to calculate a,b,c and d....becontinued....$
$l e t I = \int \frac{s i n x}{1 + s i n x + s i n (2 x)} d x \Rightarrow I = \int \frac{s i n x}{1 + s i n x + 2 s i n x c o s x} d x$ $c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int \frac{\frac{2 t}{1 + t^{2}}}{1 + \frac{2 t}{1 + t^{2}} + 2 \frac{2 t}{1 + t^{2}} \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{4 t}{{(1 + t^{2})}^{2} {1 + \frac{2 t}{1 + t^{2}} + \frac{4 t (1 - t^{2})}{{(1 + t^{2})}^{2}}}} d t$ $= \int \frac{4 t d t}{{(1 + t^{2})}^{2} + 2 t (1 + t^{2}) + 4 t (1 - t^{2})} = \int \frac{4 t d t}{t^{4} + 2 t^{2} + 1 + 2 t + 2 t^{3} + 4 t - 4 t^{3}}$ $= \int \frac{4 t d t}{t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1} l e t d e c o m p o s e t h e f r a c t i o n$ $F (t) = \frac{4 t}{t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1}$ $t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1 = 0 t h e r o o t s a r e$ $t_{1} = 1, 5898 + 1, 7445 i (c o m p l e x)$ $t_{2} \sim 1, 5898 - 1, 7445 i (c o m p l e x)$ $t_{3} = - 1 (r e a l)$ $t_{4} \sim - 0, 1795 (r e a l) \Rightarrow F (t) \sim \frac{4 t}{(t - t_{1}) (t - {\bar{t}}_{1}) (t + 1) (t - t_{4})}$ $= \frac{4 t}{(t^{2} - 2 R e (t_{1}) t + ∣ t_{1} ∣^{2}) (t + 1) (t - t_{4})} = \frac{a}{t + 1} + \frac{b}{t - t_{4}} + \frac{c t + d}{t^{2} - 2 R e (t_{1}) t + ∣ t_{1} ∣^{2}}$ $\Rightarrow I = a l n ∣ t + 1 ∣ + b l n ∣ t - t_{4} ∣ + \int \frac{c t + d}{t^{2} - 2 R e (t_{1}) t + ∣ t_{1} ∣^{2}}$ $r e s t t o c a l c u l a t e a, b, c a n d d . . . . b e c o n t i n u e d . . . .$
Commented by ~ À ® @ 237 ~ last updated on 18/Aug/19

$L e t n a m e d i t F (x)$ $W e c a n e x p l i c i t t h e d e n o m i n a t o r$ $1 + s i n x + s i n (2 x) = I m (1 + e^{i x} + e^{i 2 x}) = I m (\frac{e^{i 3 x} - 1}{e^{i x} - 1}) = I m (\frac{e^{i \frac{3 x}{2}} (e^{i \frac{3 x}{2}} - e^{- i \frac{3 x}{2}})}{e^{i \frac{x}{2}} (e^{i \frac{x}{2}} - e^{- i \frac{x}{2}})}) = I m (\frac{e^{i x} . s i n (\frac{3 x}{2})}{s i n (\frac{x}{2})}) = \frac{s i n (3 (\frac{x}{2}))}{s i n (\frac{x}{2})} I m (e^{i x}) = (3 - 4 {s i n}^{2} (\frac{x}{2})) s i n x u s i n g (s i n 3 θ = 3 s i n θ - 4 {s i n}^{3} θ)$ $N o w$ $F (x) = \int \frac{d x}{3 - 4 {s i n}^{2} (\frac{x}{2})} = 2 G (u) a v e c G (u) = \int \frac{d u}{3 - 4 {s i n}^{2} u} (s t a t i n g x = 2 u)$ $G (u) = \int \frac{\frac{1}{{c o s}^{2} u}}{\frac{3}{{c o s}^{2} u} - 4 {t a n}^{2} u} d u = \int \frac{1 + {t a n}^{2} u}{3 - {t a n}^{2} u} d u (r e m e m b e r \frac{1}{{c o s}^{2} u} = 1 + {t a n}^{2} u)$ $= \frac{1}{3} \int \frac{(1 + {t a n}^{2} u)}{1 - {(\frac{t a n u}{\sqrt{3}})}^{2}} d u = \frac{1}{\sqrt{3}} a r g t h (\frac{t a n u}{\sqrt{3}}) + c$ $F i n a l l y w e g e t$ $F (x) = \frac{2}{\sqrt{3}} a r g t h (\frac{t a n (\frac{x}{2})}{\sqrt{3}}) + k$
	Answered by Rio Michael last updated on 17/Aug/19

	$\int \frac{s i n x}{1 + s i n x + s i n 2 x} = \int s i n x d x + \int d x + \int \frac{1}{2 c o s x} d x$ $= - c o s x + x + \frac{1}{2} l n ∣ s e c x + t a n x ∣ + c$ $= x - c o s x + \frac{1}{2} l n ∣ s e c x + t a n x ∣ + c$ $p l e a s e c h e c k .$
	Commented by mr W last updated on 17/Aug/19

	$t o t a l l y w r o n g s i r!$ $\frac{a}{x + y + z} \neq \frac{a}{x} + \frac{a}{y} + \frac{a}{z}!$
	Commented by MJS last updated on 17/Aug/19
	$wrong. learn the laws of calculating with fractiond (a/(a+b))≠(a/b)+(a/c) (a/b)+(a/c)=((ac)/(bc))+((ab)/(bc))=((a(b+c))/(bc)) how could this be true: (5/6)=(5/(5+1))=(5/5)+(5/1)=1+5=6 but (5/6)=(5/(2+4))=(5/2)+(5/4)=((15)/4) and (5/6)=(5/(3+3))=(5/3)+(5/3)=((10)/3) ⇒^? (5/6)=6=((15)/4)=((10)/3)$
	$wrong . learn the laws of calculating with fractiond$ $\frac{a}{a + b} \neq \frac{a}{b} + \frac{a}{c}$ $\frac{a}{b} + \frac{a}{c} = \frac{a c}{b c} + \frac{a b}{b c} = \frac{a (b + c)}{b c}$ $how could this be true : \frac{5}{6} = \frac{5}{5 + 1} = \frac{5}{5} + \frac{5}{1} = 1 + 5 = 6 but$ $\frac{5}{6} = \frac{5}{2 + 4} = \frac{5}{2} + \frac{5}{4} = \frac{15}{4} and \frac{5}{6} = \frac{5}{3 + 3} = \frac{5}{3} + \frac{5}{3} = \frac{10}{3}$ $\overset{?}{\Rightarrow} \frac{5}{6} = 6 = \frac{15}{4} = \frac{10}{3}$
	Commented by Cmr 237 last updated on 17/Aug/19

	Answered by Cmr 237 last updated on 17/Aug/19

	Commented by MJS last updated on 17/Aug/19

	$you simply typed this into a solver . but can$ $you explain what happened along the path ?$
	Commented by mathmax by abdo last updated on 18/Aug/19

	$t h i s s o l u t i o n i s n o t c o r r e c t .$
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