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Question Number 146602 by mnjuly1970 last updated on 14/Jul/21

Answered by qaz last updated on 15/Jul/21

$$\int \frac{1-\cos \left(12\mathrm{x}\right)}{2\sin \mathrm{x}}\mathrm{dx}=\int \frac{\sin {}^2\left(6\mathrm{x}\right)}{\sin \mathrm{x}}\mathrm{dx}$$$$=4\int \frac{{\left[\sin \left(3\mathrm{x}\right)\cos \left(3\mathrm{x}\right)\right]}^2}{\sin \mathrm{x}}\mathrm{dx}$$$$=4\int \frac{{\left[(3\sin \mathrm{x}-4\sin {}^3\mathrm{x})(4\cos {}^3\mathrm{x}-3\cos \mathrm{x})\right]}^2}{\sin \mathrm{x}}\mathrm{dx}$$$$=4\int \sin \mathrm{xcos}{}^2\mathrm{x}{\left[(3-4\sin {}^2\mathrm{x})(4\cos {}^2\mathrm{x}-3)\right]}^2\mathrm{dx}$$$$=4\int {\mathrm{SC}}^2{\left[({4\mathrm{C}}^2-1)({4\mathrm{C}}^2-3)\right]}^2\mathrm{dx}$$$$=-4\int ({256\mathrm{C}}^{10}-{512\mathrm{C}}^8+{352\mathrm{C}}^6-{96\mathrm{C}}^4+{9\mathrm{C}}^2)\mathrm{dC}$$$$=-4(\frac{256}{11}{\mathrm{C}}^{11}-\frac{512}{9}{\mathrm{C}}^9+\frac{352}{7}{\mathrm{C}}^7-\frac{96}{5}{\mathrm{C}}^5+{3\mathrm{C}}^3)+\mathrm{K}$$$$=-\frac{1024}{11}\cos {}^{11}\mathrm{x}+\frac{2048}{9}\cos {}^9\mathrm{x}-\frac{1408}{7}\cos {}^7\mathrm{x}+\frac{384}{5}\cos {}^5\mathrm{x}-12\cos {}^3\mathrm{x}+\mathrm{K}$$

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