Question Number 48474 by mondodotto@gmail.com last updated on 24/Nov/18
$$\boldsymbol{\mathrm{evaluate}}\:\int_{\mathrm{0}} ^{\pi} \boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{{x}\mathrm{d}{x}} \\ $$
Answered by hassentimol last updated on 24/Nov/18
Commented by hassentimol last updated on 24/Nov/18
$$\mathrm{I}\:\mathrm{was}\:\mathrm{in}\:\mathrm{degrees}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18
$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid{x}−\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\pi \\ $$$${or}\:\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{2}} {xdx}\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:{when}\:{f}\left({x}\right)={f}\left(\mathrm{2}{a}−{x}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} {xdx} \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\mid{x}−\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}} \\ $$
Commented by mondodotto@gmail.com last updated on 26/Nov/18
$$\mathrm{sir}\:\mathrm{it}\:\mathrm{is}\:\mathrm{sin}\:\mathrm{x}^{\mathrm{2}\:} \:\mathrm{not}\:\mathrm{sin}^{\mathrm{2}} \mathrm{x} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 26/Nov/18
$${ok}…{but}\:{in}\:{question}\:{it}\:{is}\:{sin}^{\mathrm{2}} {x} \\ $$
Commented by hassentimol last updated on 26/Nov/18
$$\mathrm{Well}… \\ $$$$\mathrm{sin}\left({x}\right)^{\mathrm{2}} \:=\:\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{2}} \:=\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)\:\mathrm{by}\:\mathrm{definition} \\ $$