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Question Number 48474 by mondodotto@gmail.com last updated on 24/Nov/18

evaluate ∫_0 ^π sin^2 xdx

$$\boldsymbol{\mathrm{evaluate}}\:\int_{\mathrm{0}} ^{\pi} \boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{{x}\mathrm{d}{x}} \\ $$

Answered by hassentimol last updated on 24/Nov/18

Commented by hassentimol last updated on 24/Nov/18

I was in degrees...

$$\mathrm{I}\:\mathrm{was}\:\mathrm{in}\:\mathrm{degrees}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18

∫_0 ^π ((1−cos2x)/2)dx  =(1/2)∣x−((sin2x)/2)∣_0 ^π   =(1/2)×π  or ∫_0 ^π sin^2 xdx    ∫_0 ^(2a) f(x)dx=2∫_0 ^a f(x)dx when f(x)=f(2a−x)  =2∫_0 ^(π/2) sin^2 xdx  =2×(1/2)∣x−((sin2x)/2)∣_0 ^(π/2)   =2×(1/2)×(π/2)=(π/2)

$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−{cos}\mathrm{2}{x}}{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid{x}−\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\pi \\ $$$${or}\:\int_{\mathrm{0}} ^{\pi} {sin}^{\mathrm{2}} {xdx}\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}{a}} {f}\left({x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}\:{when}\:{f}\left({x}\right)={f}\left(\mathrm{2}{a}−{x}\right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} {xdx} \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\mid{x}−\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}} \\ $$

Commented by mondodotto@gmail.com last updated on 26/Nov/18

sir it is sin x^(2 )  not sin^2 x

$$\mathrm{sir}\:\mathrm{it}\:\mathrm{is}\:\mathrm{sin}\:\mathrm{x}^{\mathrm{2}\:} \:\mathrm{not}\:\mathrm{sin}^{\mathrm{2}} \mathrm{x} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 26/Nov/18

ok...but in question it is sin^2 x

$${ok}...{but}\:{in}\:{question}\:{it}\:{is}\:{sin}^{\mathrm{2}} {x} \\ $$

Commented by hassentimol last updated on 26/Nov/18

Well...  sin(x)^2  = (sin(x))^2  = sin^2 (x) by definition

$$\mathrm{Well}... \\ $$$$\mathrm{sin}\left({x}\right)^{\mathrm{2}} \:=\:\left(\mathrm{sin}\left({x}\right)\right)^{\mathrm{2}} \:=\:\mathrm{sin}^{\mathrm{2}} \left({x}\right)\:\mathrm{by}\:\mathrm{definition} \\ $$

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