I-dx-sinx-sin2x-

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

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“ 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 = \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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