find-dx-1-sinx-sin-2x-

Question Number 68033 by mathmax by abdo last updated on 03/Sep/19

f i n d \int \frac{d x}{1 + s i n x + s i n (2 x)}

Answered by MJS last updated on 04/Sep/19

I think I did this before

we have to use Weierstrass

\int \frac{d x}{1 + \sin x + \sin 2 x} =

[t = \tan \frac{x}{2} \to d x = \frac{2 d t}{t^{2} + 1}]

= 2 \int \frac{t^{2} + 1}{(t + 1) (t^{3} - 3 t^{2} + 5 t + 1)} d t =

= 2 \int \frac{t^{2} + 1}{(t + 1) (t - α) (t - β) (t - γ)} d t

α = 1 + u + v

β = 1 + (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) u - (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) v

γ = 1 + (- \frac{1}{2} - \frac{\sqrt{3}}{2} i) u - (- \frac{1}{2} + \frac{\sqrt{3}}{2} i) v

u = \sqrt[3]{- 2 + \frac{2}{9} \sqrt{87}}

v = \sqrt[3]{2 + \frac{2}{9} \sqrt{87}}

now we have to decompose . the path is easy

but the constants are terrible

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