Question Number 178550 by cortano1 last updated on 18/Oct/22
Answered by Ar Brandon last updated on 18/Oct/22
![I=∫_(π/(12)) ^(π/8) (((7+cos4ϑ)cos2ϑ)/(1−cos4ϑ))(((9−cos4ϑ)/(sin2ϑ)))^(2021) dϑ =∫_(π/(12)) ^(π/8) (((6+2cos^2 2ϑ)cos2ϑ)/(2sin^2 2ϑ))(((8+2sin^2 2ϑ)/(sin2ϑ)))^(2021) dϑ =∫_(π/(12)) ^(π/8) (((8−2sin^2 2ϑ)cos2ϑ)/(2sin^2 2ϑ))(8cosec2ϑ+2sin2ϑ)^(2021) dϑ =(1/2)∫_(π/(12)) ^(π/8) (8cosec2ϑcot2ϑ−2cos2ϑ)(8cosec2ϑ+2sin2ϑ)^(2021) dϑ =−(1/4)∫_(π/(12)) ^(π/8) (8cosec2ϑ+2sin2ϑ)^(2021) d(8cosec2ϑ+2sin2ϑ) =(1/(4×2022))[(8cosec2ϑ+2sin2ϑ)^(2022) ]_(π/8) ^(π/(12)) =(1/(8088))[16+1−(8(√2)+(√2))]=((17−9(√2))/(8088))](https://www.tinkutara.com/question/Q178551.png)
$${I}=\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{8}}} \frac{\left(\mathrm{7}+\mathrm{cos4}\vartheta\right)\mathrm{cos2}\vartheta}{\mathrm{1}−\mathrm{cos4}\vartheta}\left(\frac{\mathrm{9}−\mathrm{cos4}\vartheta}{\mathrm{sin2}\vartheta}\right)^{\mathrm{2021}} {d}\vartheta \\ $$$$\:\:=\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{8}}} \frac{\left(\mathrm{6}+\mathrm{2cos}^{\mathrm{2}} \mathrm{2}\vartheta\right)\mathrm{cos2}\vartheta}{\mathrm{2sin}^{\mathrm{2}} \mathrm{2}\vartheta}\left(\frac{\mathrm{8}+\mathrm{2sin}^{\mathrm{2}} \mathrm{2}\vartheta}{\mathrm{sin2}\vartheta}\right)^{\mathrm{2021}} {d}\vartheta \\ $$$$\:\:=\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{8}}} \frac{\left(\mathrm{8}−\mathrm{2sin}^{\mathrm{2}} \mathrm{2}\vartheta\right)\mathrm{cos2}\vartheta}{\mathrm{2sin}^{\mathrm{2}} \mathrm{2}\vartheta}\left(\mathrm{8cosec2}\vartheta+\mathrm{2sin2}\vartheta\right)^{\mathrm{2021}} {d}\vartheta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{8}}} \left(\mathrm{8cosec2}\vartheta\mathrm{cot2}\vartheta−\mathrm{2cos2}\vartheta\right)\left(\mathrm{8cosec2}\vartheta+\mathrm{2sin2}\vartheta\right)^{\mathrm{2021}} {d}\vartheta \\ $$$$\:\:=−\frac{\mathrm{1}}{\mathrm{4}}\int_{\frac{\pi}{\mathrm{12}}} ^{\frac{\pi}{\mathrm{8}}} \left(\mathrm{8cosec2}\vartheta+\mathrm{2sin2}\vartheta\right)^{\mathrm{2021}} {d}\left(\mathrm{8cosec2}\vartheta+\mathrm{2sin2}\vartheta\right) \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{4}×\mathrm{2022}}\left[\left(\mathrm{8cosec2}\vartheta+\mathrm{2sin2}\vartheta\right)^{\mathrm{2022}} \right]_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{12}}} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{8088}}\left[\mathrm{16}+\mathrm{1}−\left(\mathrm{8}\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}}\right)\right]=\frac{\mathrm{17}−\mathrm{9}\sqrt{\mathrm{2}}}{\mathrm{8088}} \\ $$
Commented by cortano1 last updated on 18/Oct/22
$$\mathrm{nice} \\ $$