Question Number 47638 by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18
$$\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}\:\:{dx} \\ $$
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.](Q47649.png)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}{dx}\:\Rightarrow \\ $$$${A}=\int_{\mathrm{0}} ^{\pi} \sqrt{{cos}^{\mathrm{2}} {x}}{dx}\:=\int_{\mathrm{0}} ^{\pi} \mid{cosx}\mid{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosxdx}\:−\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {cosxdx} \\ $$$$=\left[{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\left[{sinx}\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} \:=\mathrm{1}−\left(−\mathrm{1}\right)=\mathrm{2}. \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18
$${thank}\:{you}\:{sir}... \\ $$
Commented by prof Abdo imad last updated on 12/Nov/18
$${you}\:{are}\:{welcome}\:{sir}. \\ $$