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0-pi-1-cos2x-2-dx-




Question Number 47638 by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18
∫_0 ^π (√((1+cos2x)/2))  dx
$$\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}\:\:{dx} \\ $$
Commented by prof Abdo imad last updated on 12/Nov/18
let A =∫_0 ^π (√((1+cos(2x))/2))dx ⇒  A=∫_0 ^π (√(cos^2 x))dx =∫_0 ^π ∣cosx∣dx  =∫_0 ^(π/2) cosxdx −∫_(π/2) ^π cosxdx  =[sinx]_0 ^(π/2) −[sinx]_(π/2) ^π  =1−(−1)=2.
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}{dx}\:\Rightarrow \\ $$$${A}=\int_{\mathrm{0}} ^{\pi} \sqrt{{cos}^{\mathrm{2}} {x}}{dx}\:=\int_{\mathrm{0}} ^{\pi} \mid{cosx}\mid{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosxdx}\:−\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {cosxdx} \\ $$$$=\left[{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\left[{sinx}\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:=\mathrm{1}−\left(−\mathrm{1}\right)=\mathrm{2}. \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18
thank you sir...
$${thank}\:{you}\:{sir}… \\ $$
Commented by prof Abdo imad last updated on 12/Nov/18
you are welcome sir.
$${you}\:{are}\:{welcome}\:{sir}. \\ $$

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