dx-3sin-x-sin-3-x-

Question Number 113275 by bemath last updated on 12/Sep/20

\int \frac{dx}{3 \sin x + \sin^{3} x} ?

Answered by bemath last updated on 12/Sep/20

I = \int \frac{dx}{\sin x (3 + \sin^{2} x)}

I = \int \frac{\sin x dx}{\sin^{2} x (3 + \sin^{2} x)}

I = \int \frac{- d (\cos x)}{(1 - \cos^{2} x) (4 - \cos^{2} x)}

I = \int \frac{- du}{(1 - u^{2}) (4 - u^{2})}; where u = \cos x

I = - \int \frac{du}{(1 + u) (1 - u) (2 + u) (2 - u)}

I = \frac{1}{6} \int (\frac{1}{u - 1}) - (\frac{1}{u + 1}) du - \frac{1}{12} \int (\frac{1}{u - 2} - \frac{1}{u + 2}) du

I = \frac{1}{6} \ln ∣ \frac{u - 1}{u + 1} ∣ - \frac{1}{12} \ln ∣ \frac{u - 2}{u + 2} ∣ + c

I = \frac{1}{6} \ln ∣ \frac{\cos x - 1}{\cos x + 1} ∣ - \frac{1}{12} \ln ∣ \frac{\cos x - 2}{\cos x + 2} ∣ + c

Answered by abdomsup last updated on 12/Sep/20

I = \int \frac{d x}{{s i n}^{3} x + 3 s i n x} \Rightarrow

I = \int \frac{d x}{s i n x ({s i n}^{2} x + 3)}

l e t d e c o m p o s e F (u) = \frac{1}{u (u^{2} + 3)}

F (u) = \frac{a}{u} + \frac{b u + c}{u^{2} + 3}

a = \frac{1}{3}, {l i m}_{u \to + \infty} u F (u) = 0

a + b \Rightarrow b = - \frac{1}{3}

F (- u) = - F (u) \Rightarrow c = 0 \Rightarrow

F (u) = \frac{1}{3 u} - \frac{u}{3 (u^{2} + 3)}

\Rightarrow I = \frac{1}{3} \int \frac{d x}{s i n x} - \frac{1}{3} \int \frac{s i n x}{{s i n}^{2} x + 3}

\int \frac{d x}{s i n x} =_{t s n (\frac{x}{2}) = t} \int \frac{2 d t}{(1 + t^{2}) \frac{2 t}{1 + t^{2}}}

= \int \frac{d t}{t} = l n ∣ t ∣ + c_{1} = l n ∣ t a n (\frac{x}{2}) ∣ + c_{1}

\int \frac{s i n x}{{s i n}^{2} x + 3} d x = \int \frac{s i n x d x}{4 - {c o s}^{2} x}

=_{c o s x = t} \int \frac{- d t}{4 - t^{2}} = \int \frac{d t}{(t - 2) (t + 2)}

= \frac{1}{4} \int (\frac{1}{t - 2} - \frac{1}{t + 2}) d t = \frac{1}{4} l n ∣ \frac{t - 2}{t + 2} ∣ + c_{2}

= \frac{1}{4} l n ∣ \frac{c o s x - 2}{c o s x + 2} ∣ + c_{2} \Rightarrow

I = \frac{1}{3} l n ∣ t a n (\frac{x}{2}) ∣ - \frac{1}{12} l n (\frac{2 - c o s x}{2 + c o s x}) + C

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