Question Number 216618 by Tawa11 last updated on 12/Feb/25
$$\int_{\:\mathrm{0}} ^{\:\mathrm{2}\pi} \:\sqrt{\mathrm{1}\:\:−\:\:\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\:\:\mathrm{dx} \\ $$$$\mathrm{Is}\:\mathrm{the}\:\mathrm{answer}\:\:\mathrm{0}\:\:\:\mathrm{or}\:\:\:\:\mathrm{4}???? \\ $$
Answered by A5T last updated on 12/Feb/25
![∫_0 ^(2π) (√(1−cos^2 x))dx=∫_0 ^(2π) ∣sinx∣ dx ≠ ∫_0 ^(2π) sin(x) dx ∫_0 ^(2π) ∣sinx∣=∫_0 ^π sin(x) dx+∫_π ^(2π) −sin(x) dx =[−cos(x)]∣_0 ^π +[cosx]∣_π ^(2π) =−cosπ+cos0+[cos2π−cosπ] =4](Q216624.png)
$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \mathrm{x}}\mathrm{dx}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid\mathrm{sinx}\mid\:\mathrm{dx}\:\neq\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{sin}\left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid\mathrm{sinx}\mid=\int_{\mathrm{0}} ^{\pi} \mathrm{sin}\left(\mathrm{x}\right)\:\mathrm{dx}+\int_{\pi} ^{\mathrm{2}\pi} −\mathrm{sin}\left(\mathrm{x}\right)\:\mathrm{dx} \\ $$$$=\left[−\mathrm{cos}\left(\mathrm{x}\right)\right]\mid_{\mathrm{0}} ^{\pi} +\left[\mathrm{cosx}\right]\mid_{\pi} ^{\mathrm{2}\pi} \\ $$$$=−\mathrm{cos}\pi+\mathrm{cos0}+\left[\mathrm{cos2}\pi−\mathrm{cos}\pi\right] \\ $$$$=\mathrm{4} \\ $$
Commented by Tawa11 last updated on 14/Feb/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Answered by mehdee7396 last updated on 12/Feb/25
![∫_0 ^(2π) ∣sinx∣dx=∫_0 ^π sinxdx−∫_π ^(2π) sinxdx =(−cosx)]_0 ^π −(−cosx)]_π ^(2π) =(1+1)−(−1−1)=4 ✓](Q216625.png)
$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid{sinx}\mid{dx}=\int_{\mathrm{0}} ^{\pi} {sinxdx}−\int_{\pi} ^{\mathrm{2}\pi} {sinxdx} \\ $$$$\left.=\left.\left(−{cosx}\right)\right]_{\mathrm{0}} ^{\pi} −\left(−{cosx}\right)\right]_{\pi} ^{\mathrm{2}\pi} \\ $$$$=\left(\mathrm{1}+\mathrm{1}\right)−\left(−\mathrm{1}−\mathrm{1}\right)=\mathrm{4}\:\checkmark \\ $$$$ \\ $$
Commented by Tawa11 last updated on 15/Feb/25
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$