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8cos-3-x-8-sin-3-2x-dx-




Question Number 88863 by M±th+et£s last updated on 13/Apr/20
∫((8cos^3 (x))/(8+sin^3 2x))dx
$$\int\frac{\mathrm{8}{cos}^{\mathrm{3}} \left({x}\right)}{\mathrm{8}+{sin}^{\mathrm{3}} \mathrm{2}{x}}{dx} \\ $$$$ \\ $$
Commented by MJS last updated on 13/Apr/20
I tried everything I know, seems impossible  to solve  we need to factorize a polynome of 8^(th)  degree  I can factorize it to square factors but the  factors cannot be handled...
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{everything}\:\mathrm{I}\:\mathrm{know},\:\mathrm{seems}\:\mathrm{impossible} \\ $$$$\mathrm{to}\:\mathrm{solve} \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{factorize}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{8}^{\mathrm{th}} \:\mathrm{degree} \\ $$$$\mathrm{I}\:\mathrm{can}\:\mathrm{factorize}\:\mathrm{it}\:\mathrm{to}\:\mathrm{square}\:\mathrm{factors}\:\mathrm{but}\:\mathrm{the} \\ $$$$\mathrm{factors}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{handled}… \\ $$

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