Question Number 171705 by Haisokheng last updated on 20/Jun/22
Answered by puissant last updated on 20/Jun/22
![= (1/π)∫_0 ^π cos^2 (2x)dx =(1/(2π))∫_0 ^π 1+cos(4x)dx =(1/(2π))[x]_0 ^π +(1/(8π))[sin(4x)]_0 ^π = (1/2)](https://www.tinkutara.com/question/Q171711.png)
$$=\:\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} {cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\pi} \mathrm{1}+{cos}\left(\mathrm{4}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\left[{x}\right]_{\mathrm{0}} ^{\pi} +\frac{\mathrm{1}}{\mathrm{8}\pi}\left[{sin}\left(\mathrm{4}{x}\right)\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$