∫((sin x)/(1+sin x+sin 2x))dx=?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 72060 by Tushar rathore last updated on 23/Oct/19

$\int \frac{\sin x}{1 + \sin x + \sin 2 x} d x = ?$
Commented by Tushar rathore last updated on 27/Oct/19

$a n y o n e t r i e d t h i s$
Commented by mathmax by abdo last updated on 27/Oct/19
$let A =∫((sinx)/(1+sinx +sin(2x)))dx ⇒ A =∫ ((sinx)/(1+sinx +2sinx cosx))dx =_(tan((x/2))=t) ∫ (((2t)/(1+t^2 ))/(1+((2t)/(1+t^2 )) +((2(2t)(1−t^2 ))/((1+t^2 )^2 )))) ((2dt)/(1+t^2 )) = ∫ ((4t)/((1+t^2 )^2 { 1+((2t)/(1+t^2 )) +((4t(1−t^2 ))/((1+t^2 )^2 ))}))dt =4∫ (t/((1+t^2 )^2 +2t(1+t^2 ) +4t(1−t^2 )))dt =4 ∫ ((tdt)/(t^4 +2t^2 +1 +2t+2t^3 +4t−4t^3 )) =4 ∫ ((tdt)/(t^4 −2t^3 +2t^2 +6t +1)) rest decomposition of F(t) =(t/(t^4 −2t^3 +2t^2 +6t +1)) the roots of p(t)=t^4 −2t^3 +2t^2 +6t +1 are t_1 =1,5898 +1,7445i (complex) t_2 =1,5898−1,7445i(complex) t_3 =−1 (real) t_4 =−0,1795(real) ⇒F(t)=(t/((t+1)(t−t_4 )(t−t_1 )(t−t_1 ^− ))) =(t/((t+1)(t−t_4 )(t^2 −2Re(t_1 )t +∣t_1 ∣^2 ))) =(a/(t+1)) +(b/(t−t_4 )) +((ct +d)/(t^2 −2Re(t_1 )t +∣t_1 ∣^2 )) ...be continued...$
$l e t A = \int \frac{s i n x}{1 + s i n x + s i n (2 x)} d x \Rightarrow A = \int \frac{s i n x}{1 + s i n x + 2 s i n x c o s x} d x$ $=_{t a n (\frac{x}{2}) = t} \int \frac{\frac{2 t}{1 + t^{2}}}{1 + \frac{2 t}{1 + t^{2}} + \frac{2 (2 t) (1 - t^{2})}{{(1 + t^{2})}^{2}}} \frac{2 d t}{1 + t^{2}}$ $= \int \frac{4 t}{{(1 + t^{2})}^{2} {1 + \frac{2 t}{1 + t^{2}} + \frac{4 t (1 - t^{2})}{{(1 + t^{2})}^{2}}}} d t$ $= 4 \int \frac{t}{{(1 + t^{2})}^{2} + 2 t (1 + t^{2}) + 4 t (1 - t^{2})} d t$ $= 4 \int \frac{t d t}{t^{4} + 2 t^{2} + 1 + 2 t + 2 t^{3} + 4 t - 4 t^{3}}$ $= 4 \int \frac{t d t}{t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1} r e s t d e c o m p o s i t i o n o f$ $F (t) = \frac{t}{t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1} t h e r o o t s o f p (t) = t^{4} - 2 t^{3} + 2 t^{2} + 6 t + 1$ $a r e t_{1} = 1, 5898 + 1, 7445 i (c o m p l e x)$ $t_{2} = 1, 5898 - 1, 7445 i (c o m p l e x)$ $t_{3} = - 1 (r e a l)$ $t_{4} = - 0, 1795 (r e a l) \Rightarrow F (t) = \frac{t}{(t + 1) (t - t_{4}) (t - t_{1}) (t - {\bar{t}}_{1})}$ $= \frac{t}{(t + 1) (t - t_{4}) (t^{2} - 2 R e (t_{1}) t + ∣ t_{1} ∣^{2})} = \frac{a}{t + 1} + \frac{b}{t - t_{4}} + \frac{c t + d}{t^{2} - 2 R e (t_{1}) t + ∣ t_{1} ∣^{2}}$ $. . . b e c o n t i n u e d . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum