Question Number 226668 by Spillover last updated on 09/Dec/25
Answered by Frix last updated on 09/Dec/25
$$=\underset{\mathrm{0}} {\overset{\pi} {\int}}\left(\mathrm{2cos}\:\mathrm{16}{x}\:+\mathrm{4cos}\:\mathrm{14}{x}\:+\mathrm{2cos}\:\mathrm{12}{x}\:−\mathrm{4cos}\:\mathrm{10}{x}\:−\mathrm{8cos}\:\mathrm{8}{x}\:−\mathrm{8cos}\:\mathrm{6}{x}\:−\mathrm{2cos}\:\mathrm{4}{x}\:+\mathrm{8cos}\:\mathrm{2}{x}\:+\mathrm{7}\right){dx}= \\ $$$$… \\ $$$$=\mathrm{7}\pi \\ $$