Evaluate-sinx-x-4-x-2-1-dx-I-need-full-detailed-explanation-thank-you-in-advance-

Question Number 204804 by Mastermind last updated on 27/Feb/24

Evaluate \int \frac{sinx}{x^{4} + x^{2} + 1} dx

I need full detailed explanation, thank you in

advance .

Commented by TonyCWX08 last updated on 28/Feb/24

I w o u l d u s e t h e C o n t o u r I n t e g r a t i o n b u t t h e t e i s n o i n t e r v a l \dots

Commented by Mastermind last updated on 28/Feb/24

Kindly evaluate \dots or what do you think we

can do ?

Commented by TonyCWX08 last updated on 28/Feb/24

C o n t o u r I n t e g r a l i s u s e d i n C o m p l e x W o r l d

U s u a l l y u s e d f o r t h e i n t e g r a l w i t h i n t e r v a l (- \infty, \infty)

B u t i n t h i s c a s e, i t {c a n}^{'} t b e u s e d .

Answered by Frix last updated on 28/Feb/24

\frac{\sin x}{x^{4} + x^{2} + 1} = \frac{\sin x}{(x^{2} - x + 1) (x^{2} + x + 1)} =

= \frac{(1 - x) \sin x}{x^{2} - x + 1} + \frac{(1 + x) \sin x}{x^{2} + x + 1} =

= \frac{\sin x}{x^{2} - x + 1} - \frac{x \sin x}{x^{2} - x + 1} + \frac{\sin x}{x^{2} + x + 1} + \frac{x \sin x}{x^{2} + x + 1} =

= \frac{4 \sin x}{(2 x - 1 - \sqrt{3} i) (2 x - 1 + \sqrt{3} i)} -

- \frac{4 x \sin x}{(2 x - 1 - \sqrt{3} i) (2 x - 1 + \sqrt{3} i)} +

+ \frac{4 \sin x}{(2 x + 1 - \sqrt{3} i) (2 x + 1 + \sqrt{3} i)} +

+ \frac{4 x \sin x}{(2 x + 1 - \sqrt{3} i) (2 x + 1 + \sqrt{3} i)} =

= \frac{2 isin x}{\sqrt{3} (2 x - 1 + \sqrt{3} i)} - \frac{2 isin x}{\sqrt{3} (2 x - 1 - \sqrt{3} i)} +

+ \frac{(1 - \sqrt{3} i) \sin x}{\sqrt{3} (2 x - 1 - \sqrt{3} i)} - \frac{(\sqrt{3} + i) \sin x}{\sqrt{3} (2 x - 1 + \sqrt{3} i)} -

- \frac{2 isin x}{\sqrt{3} (2 x + 1 - \sqrt{3} i)} + \frac{2 isin x}{\sqrt{3} (2 x + 1 + \sqrt{3} i)} +

+ \frac{(\sqrt{3} + i) \sin x}{\sqrt{3} (2 x + 1 - \sqrt{3} i)} + \frac{(\sqrt{3} - i) \sin x}{\sqrt{3} (2 x + 1 + \sqrt{3} i)}

(Please check for sign errors)

So essentially we need to solve integrals

of the shape

a \int \frac{\sin x}{b x + c} d x = \frac{a}{b} (\cos \frac{c}{b} Si (x + \frac{c}{b}) - \sin \frac{c}{b} Ci (x + \frac{c}{b})) + C

with Si (Ci) being the Integral (Co) Sine

\int \frac{\sin x}{x} d x = Si x + C \int \frac{\cos x}{x} d x = Ci x + C

Somebody finish this please .

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