Question Number 39834 by math khazana by abdo last updated on 12/Jul/18
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\mid\:{cos}\left(\mathrm{2}{x}\right)−{cos}\left(\mathrm{3}{x}\right)\mid{dx} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18
![∫_0 ^(Π/6) (cos2x−cos3x)dx reason value of cos2x>cos3x in [0,(Π/6)] =∣((sin2x)/2)−((sin3x)/3)∣_0 ^(Π/6) =(((sin(Π/3))/2)−((sin(Π/2))/3))=(((√3) )/4)−(1/3)=((3(√3) −4)/(12))](https://www.tinkutara.com/question/Q39841.png)
$$\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{6}}} \left({cos}\mathrm{2}{x}−{cos}\mathrm{3}{x}\right){dx} \\ $$$${reason}\:{value}\:{of}\:{cos}\mathrm{2}{x}>{cos}\mathrm{3}{x}\:{in}\:\left[\mathrm{0},\frac{\Pi}{\mathrm{6}}\right] \\ $$$$=\mid\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}−\frac{{sin}\mathrm{3}{x}}{\mathrm{3}}\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{6}}} \\ $$$$=\left(\frac{{sin}\frac{\Pi}{\mathrm{3}}}{\mathrm{2}}−\frac{{sin}\frac{\Pi}{\mathrm{2}}}{\mathrm{3}}\right)=\frac{\sqrt{\mathrm{3}}\:}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{4}}{\mathrm{12}} \\ $$