Question Number 196940 by sonukgindia last updated on 04/Sep/23
Answered by MM42 last updated on 04/Sep/23
$${sin}\mathrm{8}{x}=\mathrm{8}{sinxcosxcos}\mathrm{2}{xcos}\mathrm{4}{x} \\ $$$$\Rightarrow{I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {cosxcos}\mathrm{2}{xcos}\mathrm{4}{xdx} \\ $$$${cosxcos}\mathrm{2}{xcos}\mathrm{4}{x}=\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{7}{x}+{cos}\mathrm{5}{x}+{cos}\mathrm{3}{x}+{cosx}\right) \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{7}}{sin}\mathrm{7}{x}+\frac{\mathrm{1}}{\mathrm{5}}{sin}\mathrm{5}{x}+\frac{\mathrm{1}}{\mathrm{3}}{sin}\mathrm{3}{x}+{sinx}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(−\frac{\mathrm{1}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{1}\right)=\frac{\mathrm{26}\sqrt{\mathrm{2}}}{\mathrm{105}}\:\checkmark\: \\ $$