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Question Number 178550 by cortano1 last updated on 18/Oct/22

Answered by Ar Brandon last updated on 18/Oct/22

$$I={\int }_{\frac{\pi }{12}}^{\frac{\pi }{8}}\frac{(7+\cos 4\vartheta )\cos 2\vartheta }{1-\cos 4\vartheta }{\left(\frac{9-\cos 4\vartheta }{\sin 2\vartheta }\right)}^{2021}d\vartheta$$$$={\int }_{\frac{\pi }{12}}^{\frac{\pi }{8}}\frac{(6+{2\cos }^{2}2\vartheta )\cos 2\vartheta }{{2\sin }^{2}2\vartheta }{\left(\frac{8+{2\sin }^{2}2\vartheta }{\sin 2\vartheta }\right)}^{2021}d\vartheta$$$$={\int }_{\frac{\pi }{12}}^{\frac{\pi }{8}}\frac{(8-{2\sin }^{2}2\vartheta )\cos 2\vartheta }{{2\sin }^{2}2\vartheta }{(8\mathrm{cosec}2\vartheta +2\sin 2\vartheta )}^{2021}d\vartheta$$$$=\frac{1}{2}{\int }_{\frac{\pi }{12}}^{\frac{\pi }{8}}(8\mathrm{cosec}2\vartheta \cot 2\vartheta -2\cos 2\vartheta ){(8\mathrm{cosec}2\vartheta +2\sin 2\vartheta )}^{2021}d\vartheta$$$$=-\frac{1}{4}{\int }_{\frac{\pi }{12}}^{\frac{\pi }{8}}{(8\mathrm{cosec}2\vartheta +2\sin 2\vartheta )}^{2021}d(8\mathrm{cosec}2\vartheta +2\sin 2\vartheta )$$$$=\frac{1}{4\times 2022}{\left[{(8\mathrm{cosec}2\vartheta +2\sin 2\vartheta )}^{2022}\right]}_{\frac{\pi }{8}}^{\frac{\pi }{12}}$$$$=\frac{1}{8088}[16+1-(8\sqrt{2}+\sqrt{2})]=\frac{17-9\sqrt{2}}{8088}$$

Commented by cortano1 last updated on 18/Oct/22

$$\mathrm{nice}$$

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