P ((cos x)^3)/(x^6+1)

[kinu]

\(\int\frac{\cos^3 x}{x^6+1}dx\)

[João P. Ferreira]

Hi, just a hint:

Separate \(cos^3(x)\) in terms of sums of \(cos(x)\) and \(cos(3x)\)

Then remeber that:

\(x^6+1 = (x^2+1)(x^2-1+\sqrt[3]{-1})(x^2-\sqrt[3]{-1})\)

Then you may use the functions Ci(z) and Si(z) for each term

I hope it will help

[kinu]

Thanks Sir

\(x^6+1 = (x^2+1).(x^4-x^2+1) = (x^2+1).(x^2+\sqrt{3}x+1).(x^2-\sqrt{3}x+1)\)

So \(\int\frac{\cos^3 x}{(x^2+1)(x^2+\sqrt{3}x+1).(x^2-\sqrt{3}x+1)}dx\)

Sir after that how can i solve it

thanks

[João P. Ferreira]

Remember that \(cos^3(x)=\frac{1}{4}\left(cos(3x)+3cos(x)\right)\)

Then

\(\int\frac{\cos^3 x}{(x^2+1)(x^2+\sqrt{3}x+1).(x^2-\sqrt{3}x+1)}dx=\frac{1}{4}\int\frac{\cos{3x}}{(x^2+1)(x^2+\sqrt{3}x+1).(x^2-\sqrt{3}x+1)}dx+\frac{3}{4}\int\frac{\cos{x}}{(x^2+1)(x^2+\sqrt{3}x+1).(x^2-\sqrt{3}x+1)}dx\)

Then you need to factorize \(\frac{1}{(x^2+1)(x^2+\sqrt{3}x+1).(x^2-\sqrt{3}x+1)}\) in partial fractions

\(\frac{1}{(x^2+1)(x^2+\sqrt{3}x+1).(x^2-\sqrt{3}x+1)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+\sqrt{3}x+1}+\frac{Ex+F}{x^2-\sqrt{3}x+1}\)

Then you shall use the Ci(z) and Si(z) functions for each fraction

[kinu]

Thanks Sir