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Question Number 203434 by sonukgindia last updated on 19/Jan/24

Answered by witcher3 last updated on 20/Jan/24

$$=\int \frac{-2\sin \left(2\mathrm{x}\right)\sin \left(3\mathrm{x}\right)}{1+2\cos \left(3\mathrm{x}\right)}\mathrm{dx}$$$$=\frac{\sin \left(2\mathrm{x}\right)}{3}\ln (1+2\cos \left(3\mathrm{x}\right))-\frac{1}{6}\int \sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)\ln (1+2\cos \left(3\mathrm{x}\right))\mathrm{dx}$$$$\ln (1+2\cos \left(3\mathrm{x}\right))=\ln (1+2\cos \left(\mathrm{x}\right)({2\cos }^2\left(\mathrm{x}\right)-1)-4(1-{\cos }^2\left(\mathrm{x}\right))\cos \left(\mathrm{x}\right))$$$$=\ln (1+{8\cos }^3\left(\mathrm{x}\right)-6\cos \left(\mathrm{x}\right))$$$$\cos \left(\mathrm{x}\right)=\mathrm{u}$$$$\int$$$$=\frac{1}{6}\int \mathrm{uln}(1+{8\mathrm{u}}^3-6\mathrm{u})\mathrm{du}$$$$1+{8\mathrm{u}}^3-6\mathrm{u}=8\prod _{\mathrm{w}\mid 1+{8\mathrm{w}}^3-6\mathrm{w}=0}(\mathrm{u}-\mathrm{w})$$$$=\frac{1}{6}\int \mathrm{uln}\left(8\right)+\sum _{\mathrm{w}}\mathrm{uln}(\mathrm{u}-\mathrm{w})\mathrm{du}$$$$\int \mathrm{uln}(\mathrm{u}-\mathrm{w})=\int (\mathrm{u}-\mathrm{w})\ln (\mathrm{u}-\mathrm{w})+\mathrm{wln}(\mathrm{u}-\mathrm{w})\mathrm{du}$$$$=\frac{1}{2}{(\mathrm{u}-\mathrm{w})}^2\ln (\mathrm{u}-\mathrm{w})-\frac{1}{4}(\mathrm{u}-\mathrm{w})$$$$\frac{\sin \left(2\mathrm{x}\right)}{3}\ln (1+2\cos \left(3\mathrm{x}\right))+\frac{1}{4}{\cos }^2\left(\mathrm{x}\right)\ln \left(2\right)+\frac{1}{6}\sum _{\mathrm{w}\mid {8\mathrm{w}}^3-6\mathrm{w}+1=0}(\cos \left(\mathrm{x}\right)-\mathrm{w})\ln (\cos \left(\mathrm{x}\right)-\mathrm{w})-\frac{1}{4}(\cos \left(\mathrm{x}\right)-\mathrm{w})$$$$\mathrm{w}\mathrm{can}\mathrm{bee}\mathrm{expressed}\mathrm{withe}\mathrm{cardon}$$

Answered by MathematicalUser2357 last updated on 30/Jan/24

$$\frac{\sin 2x}{3}\ln (1+2\cos 3x)+\frac{1}{4}{\cos }^{2}x\ln 2+\frac{1}{6}\sum _{8w^3-6w+1=0}((\cos x-w)\ln (\cos x-w)-\frac{1}{4}(\cos x-w))+C$$

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