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Question Number 82954 by john santu last updated on 26/Feb/20

$$\int \frac{\cos 4\mathrm{x}-\cos 2\mathrm{x}}{\sin 4\mathrm{x}-\cos 2\mathrm{x}}\mathrm{dx}$$

Commented by john santu last updated on 26/Feb/20

$$\mathrm{let}\mathrm{u}=\cos 2\mathrm{x}\Rightarrow \mathrm{dx}=-\frac{\mathrm{du}}{2\sqrt{1-{\mathrm{u}}^2}}$$$$\Rightarrow \int \frac{{2\mathrm{u}}^2-\mathrm{u}-1}{2\mathrm{u}\sqrt{1-{\mathrm{u}}^2}-\mathrm{u}}\times \frac{\mathrm{du}}{-2\sqrt{1-{\mathrm{u}}^2}}$$$$\Rightarrow \int \frac{{2\mathrm{u}}^2-\mathrm{u}-1}{-4\mathrm{u}(1-{\mathrm{u}}^2)+2\mathrm{u}\sqrt{1-{\mathrm{u}}^2}}\mathrm{du}$$$$\mathrm{stuck}$$$$$$

Commented by MJS last updated on 26/Feb/20

$$\mathrm{always}\mathrm{try}t=\tan \frac{x}{2}\iff x=2\arctan t\mathrm{but}\mathrm{in}$$$$\mathrm{this}\mathrm{case}\mathrm{we}\mathrm{have}\mathrm{only}2nx{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{enough}\mathrm{to}$$$$\mathrm{let}t=\tan x$$$$\mathrm{Weierstrass}\mathrm{substitution}:$$$$t=\tan \frac{x}{2}\to dx=\frac{2dt}{1+t^2};\sin x=\frac{2t}{1+t^2};\cos x=\frac{1-t^2}{1+t^2}$$

Commented by john santu last updated on 26/Feb/20

$$\mathrm{o}\mathrm{yes}\mathrm{sir}.{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{forgot}\mathrm{this}\mathrm{method}.$$$$\mathrm{thank}\mathrm{you}\mathrm{mister}\mathrm{mjs}$$

Answered by MJS last updated on 26/Feb/20

$$\int \frac{\cos 4x-\cos 2x}{\sin 4x-\cos 2x}dx=$$$$[t=\tan x\to dx=\frac{dt}{t^2+1}]$$$$=2\int \frac{(t^2-3)t^2}{(t-1)(t+1)(t-2-\sqrt{3})(t-2+\sqrt{3})(t^2+1)}dt=$$$$=\frac{1}{2}\int \frac{dt}{t-1}+\frac{1}{6}\int \frac{dt}{t+1}+\frac{1+\sqrt{3}}{6}\int \frac{dt}{t-2-\sqrt{3}}+\frac{1-\sqrt{3}}{6}\int \frac{dt}{t-2+\sqrt{3}}-\int \frac{t}{t^2+1}dt$$$$\mathrm{and}\mathrm{all}\mathrm{of}\mathrm{them}\mathrm{are}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

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