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Question Number 133911 by liberty last updated on 25/Feb/21

$$\mathcal{A}=\int \frac{\cos \mathrm{x}}{\sin 5\mathrm{x}+\sin \mathrm{x}}\mathrm{dx}?$$

Answered by Ar Brandon last updated on 25/Feb/21

$$\mathcal{A}=\int \frac{\mathrm{cosx}}{\sin 5\mathrm{x}+\mathrm{sinx}}\mathrm{dx}=\int \frac{\mathrm{cosx}}{2\sin 3\mathrm{xcos}2\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{2}\int \frac{\mathrm{cosx}\mathrm{dx}}{(3\mathrm{sinx}-{4\sin }^3\mathrm{x})(1-{2\sin }^2\mathrm{x})}$$$$=\frac{1}{2}\int \frac{\mathrm{du}}{(3\mathrm{u}-{4\mathrm{u}}^3)(1-{2\mathrm{u}}^2)}$$$$=\frac{1}{6}\int \frac{\mathrm{du}}{\mathrm{u}(1-\frac{2\mathrm{u}}{\sqrt{3}})(1+\frac{2\mathrm{u}}{\sqrt{3}})(1-\sqrt{2}\mathrm{u})(1+\sqrt{2}\mathrm{u})}$$$$=\frac{1}{6}\int \{\frac{1}{\mathrm{u}}-\frac{\sqrt{3}/2}{(1-2\mathrm{u}/\sqrt{3})}+\frac{\sqrt{3}/2}{(1+2\mathrm{u}/\sqrt{3})}+\frac{\sqrt{2}/3}{1-\sqrt{2}\mathrm{u}}-\frac{\sqrt{2}/3}{1+\sqrt{2}\mathrm{u}}\}\mathrm{du}$$$$=\frac{1}{6}\{\ln \mid \mathrm{sinx}\mid -\frac{3}{4}\ln \mid \frac{\sqrt{3}-2\mathrm{sinx}}{\sqrt{3}}\mid +\frac{3}{4}\ln \mid \frac{\sqrt{3}+2\mathrm{sinx}}{\sqrt{3}}\mid +\frac{\ln \mid 1-\sqrt{2}\mathrm{sinx}\mid }{3}-\frac{\ln \mid 1+\sqrt{2}\mathrm{sinx}\mid }{3}+\mathrm{C}\}$$

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