sinx-1-sinx-sin2x-dx-

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 \dots

I found no easier path

Commented by MJS last updated on 17/Aug/19

\dots 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 \dots 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

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 \dots . b e c o n t i n u e d \dots .

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

\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 .