I = ∫(dx/(sinx + sin2x))


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Question Number 6444 by sanusihammed last updated on 27/Jun/16

$I = \int \frac{d x}{s i n x + s i n 2 x}$
Commented by Temp last updated on 27/Jun/16

$Try www . WolframAlpha . com$ $search :$ $‘ ‘ i n t 1 / (s i n (x) + s i n (2 x)) {d x}^{″}$ $if you want, for example, \int_{1}^{2} x d x :$ $‘ ‘ i n t x d x f r o m x = 1 t o 2^{″}$
	Answered by Yozzii last updated on 27/Jun/16
	$I=∫(dx/(sinx+sin2x))=∫(dx/(sinx(1+2cosx))) Let u=tan0.5x⇒(2/(1+u^2 ))du=dx sinx=((2u)/(1+u^2 )) and cosx=((1−u^2 )/(1+u^2 )). ∴I=∫((2/(1+u^2 ))/(((2u)/(1+u^2 ))(1+((2(1−u^2 ))/(1+u^2 )))))du I=∫(((1+u^2 ))/(u(1+u^2 +2−2u^2 )))du I=∫((1+u^2 )/(u(3−u^2 )))du I=−∫((1+u^2 )/(u(u−(√3))(u+(√3))))du Let ((1+u^2 )/(u(u−(√3))(u+(√3))))≡(a/u)+(b/(u−(√3)))+(c/(u+(√3))) ≡((a(u^2 −3)+bu(u+(√3))+cu(u−(√3)))/(u(u+(√3))(u−(√3)))) ∴1+u^2 =a(u^2 −3)+bu(u+(√3))+cu(u−(√3)) Let u=0∴ 1=a(−3)⇒a=((−1)/3). Let u=(√3)⇒4=b(√3)(2(√3)) 4=6b⇒b=(2/3) Let u=−(√3)⇒4=c(−(√3))(−2(√3)) c=(2/3). ∴ I=−∫{(2/3)((1/(u+(√3)))+(1/(u−(√3))))−(1/(3u))}du I=−(1/3)∫{2((1/(u+(√3)))+(1/(u−(√3))))−(1/u)}du I=((−1)/3)[2ln∣(u+(√3))(u−(√3))∣−ln∣u∣]+C I=((−1)/3)ln∣(((u^2 −3)^2 )/u)∣+C I=(1/3)ln∣(u/((u^2 −3)^2 ))∣+C I=(1/3)ln∣((tan0.5x)/((tan^2 0.5x−3)^2 ))∣+C$
	$I = \int \frac{d x}{s i n x + s i n 2 x} = \int \frac{d x}{s i n x (1 + 2 c o s x)}$ $L e t u = t a n 0.5 x \Rightarrow \frac{2}{1 + u^{2}} d u = d x$ $s i n x = \frac{2 u}{1 + u^{2}} a n d c o s x = \frac{1 - u^{2}}{1 + u^{2}} .$ $∴ I = \int \frac{2 / (1 + u^{2})}{\frac{2 u}{1 + u^{2}} (1 + \frac{2 (1 - u^{2})}{1 + u^{2}})} d u$ $I = \int \frac{(1 + u^{2})}{u (1 + u^{2} + 2 - 2 u^{2})} d u$ $I = \int \frac{1 + u^{2}}{u (3 - u^{2})} d u$ $I = - \int \frac{1 + u^{2}}{u (u - \sqrt{3}) (u + \sqrt{3})} d u$ $L e t \frac{1 + u^{2}}{u (u - \sqrt{3}) (u + \sqrt{3})} \equiv \frac{a}{u} + \frac{b}{u - \sqrt{3}} + \frac{c}{u + \sqrt{3}}$ $\equiv \frac{a (u^{2} - 3) + b u (u + \sqrt{3}) + c u (u - \sqrt{3})}{u (u + \sqrt{3}) (u - \sqrt{3})}$ $∴ 1 + u^{2} = a (u^{2} - 3) + b u (u + \sqrt{3}) + c u (u - \sqrt{3})$ $L e t u = 0 ∴ 1 = a (- 3) \Rightarrow a = \frac{- 1}{3} .$ $L e t u = \sqrt{3} \Rightarrow 4 = b \sqrt{3} (2 \sqrt{3})$ $4 = 6 b \Rightarrow b = \frac{2}{3}$ $L e t u = - \sqrt{3} \Rightarrow 4 = c (- \sqrt{3}) (- 2 \sqrt{3})$ $c = \frac{2}{3} .$ $∴ I = - \int {\frac{2}{3} (\frac{1}{u + \sqrt{3}} + \frac{1}{u - \sqrt{3}}) - \frac{1}{3 u}} d u$ $I = - \frac{1}{3} \int {2 (\frac{1}{u + \sqrt{3}} + \frac{1}{u - \sqrt{3}}) - \frac{1}{u}} d u$ $I = \frac{- 1}{3} [2 l n ∣ (u + \sqrt{3}) (u - \sqrt{3}) ∣ - l n ∣ u ∣] + C$ $I = \frac{- 1}{3} l n ∣ \frac{{(u^{2} - 3)}^{2}}{u} ∣ + C$ $I = \frac{1}{3} l n ∣ \frac{u}{{(u^{2} - 3)}^{2}} ∣ + C$ $I = \frac{1}{3} l n ∣ \frac{t a n 0.5 x}{{({t a n}^{2} 0.5 x - 3)}^{2}} ∣ + C$
	Commented by nburiburu last updated on 27/Jun/16

	$t h e r e i s a n o t h e r w a y, i n s t e a d u n i v e r s a l t r i g o n o m e t r y s u b s t i t u t i o n :$ $\int \frac{1}{s i n x (1 + 2 c o s x)} . d x . \frac{s i n x}{s i n x} =$ $\int \frac{s i n x}{{s i n}^{2} x (1 + 2 c o s x)} . d x =$ $\int \frac{s i n x}{(1 - {c o s}^{2} x) (1 + 2 c o s x)} . d x$ $t = c o s x$ $d t = - s i n x d x$ $\int \frac{- 1}{(1 - t^{2}) (1 + 2 t)} . d t$ $a n d u s i n g r a t i o n a l d e s c o m p o s i t i o n i t i s s o l v e d .$ $1 - t^{2} = (1 - t) (1 + t)$ $= - \int \frac{A}{1 - t} d t - \int \frac{B}{1 + t} d t - \int \frac{C}{1 + 2 t} d t =$ $= A l n (1 - t) - B l n (1 + t) - \frac{C}{2} l n (1 + 2 t) + c$ $w h e r e$ $1 = A (1 + t) (1 + 2 t) + B (1 - t) (1 + 2 t) + C (1 - t^{2})$ $a n d i f t = 1 : 1 = A . 2.3 \Rightarrow A = \frac{1}{6}$ $i f t = - 1 : 1 = B (- 2) (- 1) \Rightarrow B = \frac{1}{2}$ $i f t = - \frac{1}{2} : 1 = C (1 - \frac{1}{4}) \Rightarrow C = \frac{4}{3}$ $s o$ $= \frac{1}{6} l n (1 - c o s x) - \frac{1}{2} l n (1 + c o s x) - \frac{2}{3} l n (1 + 2 c o s x) + c$
	Commented by sanusihammed last updated on 27/Jun/16

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