Question Number 164672 by mathlove last updated on 20/Jan/22
Answered by Ar Brandon last updated on 20/Jan/22
![Ω=∫_0 ^π ((cosx+1)/( (√(3+4cosx+cos^2 x))))dx=∫_0 ^π ((cosx+1)/( (√((cosx+1)(cosx+3)))))dx =∫_0 ^π (√((cosx+1)/(cosx+3)))dx=(√2)∫_0 ^π ((cos((x/2)))/( (√(4−2sin^2 ((x/2))))))dx =2∫_0 ^(π/2) ((cosx)/( (√(2−sin^2 x))))dx=2[arcsin(((sinx)/( (√2))))]_0 ^(π/2) =2((π/4))=(π/2)](https://www.tinkutara.com/question/Q164700.png)
$$\Omega=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}+\mathrm{4cos}{x}+\mathrm{cos}^{\mathrm{2}} {x}}}{dx}=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}{x}+\mathrm{1}}{\:\sqrt{\left(\mathrm{cos}{x}+\mathrm{1}\right)\left(\mathrm{cos}{x}+\mathrm{3}\right)}}{dx} \\ $$$$\:\:\:\:=\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{cos}{x}+\mathrm{1}}{\mathrm{cos}{x}+\mathrm{3}}}{dx}=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{cos}\left(\frac{{x}}{\mathrm{2}}\right)}{\:\sqrt{\mathrm{4}−\mathrm{2sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}{dx} \\ $$$$\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}{x}}{\:\sqrt{\mathrm{2}−\mathrm{sin}^{\mathrm{2}} {x}}}{dx}=\mathrm{2}\left[\mathrm{arcsin}\left(\frac{\mathrm{sin}{x}}{\:\sqrt{\mathrm{2}}}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{2}\left(\frac{\pi}{\mathrm{4}}\right)=\frac{\pi}{\mathrm{2}} \\ $$