0-pi-2-sin-2-4-sin-2-d-

Question Number 205558 by universe last updated on 24/Mar/24

$$\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{\mathrm{sin}^{\mathrm{2}} \mathrm{4}\theta\:}{\mathrm{sin}^{\mathrm{2}} \theta\:}{d}\theta\:\:=\:\:\:? \\ $$

Answered by Berbere last updated on 24/Mar/24

$${sin}^{\mathrm{2}} \left(\mathrm{4}{x}\right)=\mathrm{4}{sin}^{\mathrm{2}} \left(\mathrm{2}{x}\right){cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right) \\ $$$$=\mathrm{16}{sin}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{8}{cos}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{4}\left({cos}\left(\mathrm{2}{x}\right)+\mathrm{1}\right)\left({cos}\left(\mathrm{4}{x}\right)+\mathrm{1}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{4}{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right)+\mathrm{4}{cis}\left(\mathrm{2}{x}\right)+\mathrm{4}{cos}\left(\mathrm{4}{x}\right)+\mathrm{4}\:{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{6}{cos}\left(\mathrm{2}{x}\right)+\mathrm{2}{cos}\left(\mathrm{6}{x}\right)+\mathrm{4}{cos}\left(\mathrm{4}{x}\right)+\mathrm{4}{dx} \\ $$$$=\mathrm{2}\pi \\ $$$$ \\ $$

Commented by Frix last updated on 24/Mar/24

$$\frac{\mathrm{sin}^{\mathrm{2}} \:\mathrm{4}{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}=\mathrm{2cos}\:\mathrm{6}{x}\:+\mathrm{4cos}\:\mathrm{4}{x}\:+\mathrm{6cos}\:\mathrm{2}{x}\:+\mathrm{4} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{sin}^{\mathrm{2}} \:\mathrm{4}{x}}{\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:\mathrm{6}{x}\:+\mathrm{sin}\:\mathrm{4}{x}\:+\mathrm{3sin}\:\mathrm{2}{x}\:+\mathrm{4}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{2}\pi \\ $$

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