Question Number 192916 by Mingma last updated on 31/May/23
Answered by ARUNG_Brandon_MBU last updated on 31/May/23
![I=∫_0 ^(π/2) (((1−sin2x)/(1+cosx))+((1−cos2x)/(1+sinx)))dx =∫_0 ^(π/2) (dx/(1+cosx))+2∫_0 ^(π/2) ((cosx)/(1+cosx))(sinxdx)+2∫_0 ^(π/2) ((sin^2 x)/(1+sinx))dx =∫_0 ^(π/2) ((1−cosx)/(sin^2 x))dx+2∫_0 ^1 (u/(1+u))du+2∫_0 ^(π/2) (sinx−1+(1/(1+sinx)))dx =[cotx−(1/(sinx))]_(π/2) ^0 +2[u−ln(1+u)]_0 ^1 +2[cosx+x−tanx+(1/(cosx))]_(π/2) ^0 =1+2(1−ln2)+2(2−(π/2))=7−π−2ln2](https://www.tinkutara.com/question/Q192934.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}−\mathrm{sin2}{x}}{\mathrm{1}+\mathrm{cos}{x}}+\frac{\mathrm{1}−\mathrm{cos2}{x}}{\mathrm{1}+\mathrm{sin}{x}}\right){dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\mathrm{cos}{x}}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}{x}}{\mathrm{1}+\mathrm{cos}{x}}\left(\mathrm{sin}{xdx}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}{x}}{dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−\mathrm{cos}{x}}{\mathrm{sin}^{\mathrm{2}} {x}}{dx}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}}{\mathrm{1}+{u}}{du}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}{x}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}{x}}\right){dx} \\ $$$$\:\:\:=\left[\mathrm{cot}{x}−\frac{\mathrm{1}}{\mathrm{sin}{x}}\right]_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} +\mathrm{2}\left[{u}−\mathrm{ln}\left(\mathrm{1}+{u}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\left[\mathrm{cos}{x}+{x}−\mathrm{tan}{x}+\frac{\mathrm{1}}{\mathrm{cos}{x}}\right]_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \\ $$$$\:\:\:=\mathrm{1}+\mathrm{2}\left(\mathrm{1}−\mathrm{ln2}\right)+\mathrm{2}\left(\mathrm{2}−\frac{\pi}{\mathrm{2}}\right)=\mathrm{7}−\pi−\mathrm{2ln2} \\ $$
Commented by aba last updated on 31/May/23
$$\mathrm{Error}\:\mathrm{I}=\mathrm{3}−\pi+\mathrm{2ln2} \\ $$
Commented by aba last updated on 31/May/23
![I=∫_0 ^(π/2) (((1−sin2x)/(1+cosx))+((1−cos2x)/(1+sinx)))dx =∫_0 ^(π/2) (dx/(1+cosx))−2∫_0 ^(π/2) ((cosx)/(1+cosx))(sinxdx)+2∫_0 ^(π/2) ((sin^2 x)/(1+sinx))dx =∫_0 ^(π/2) ((1−cosx)/(sin^2 x))dx−2∫_0 ^1 (u/(1+u))du+2∫_0 ^(π/2) (sinx−1+(1/(1+sinx)))dx =[+cotx−(1/(sinx))]_(π/2) ^0 −2[u−ln(1+u)]_0 ^1 +2[cosx−x+tanx−(1/(cosx))]_(π/2) ^0 =1−2(1−ln2)+2(2−(π/2))=3−π+2ln2](https://www.tinkutara.com/question/Q192955.png)
$$ \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\mathrm{1}−\mathrm{sin2}{x}}{\mathrm{1}+\mathrm{cos}{x}}+\frac{\mathrm{1}−\mathrm{cos2}{x}}{\mathrm{1}+\mathrm{sin}{x}}\right){dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\mathrm{cos}{x}}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{cos}{x}}{\mathrm{1}+\mathrm{cos}{x}}\left(\mathrm{sin}{xdx}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{sin}{x}}{dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−\mathrm{cos}{x}}{\mathrm{sin}^{\mathrm{2}} {x}}{dx}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}}{\mathrm{1}+{u}}{du}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{sin}{x}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}{x}}\right){dx} \\ $$$$\:\:\:=\left[+\mathrm{cot}{x}−\frac{\mathrm{1}}{\mathrm{sin}{x}}\right]_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} −\mathrm{2}\left[{u}−\mathrm{ln}\left(\mathrm{1}+{u}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\left[\mathrm{cos}{x}−{x}+\mathrm{tanx}−\frac{\mathrm{1}}{\mathrm{cosx}}\right]_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{0}} \\ $$$$ \\ $$$$\:\:\:=\mathrm{1}−\mathrm{2}\left(\mathrm{1}−\mathrm{ln2}\right)+\mathrm{2}\left(\mathrm{2}−\frac{\pi}{\mathrm{2}}\right)=\mathrm{3}−\pi+\mathrm{2ln2} \\ $$
Commented by Mingma last updated on 01/Jun/23
Perfect ��
Commented by ARUNG_Brandon_MBU last updated on 04/Jun/23
Yeah you're right. Thanks!
Answered by aba last updated on 31/May/23
$$\mathrm{a}=\mathrm{3}\:\wedge\:\mathrm{b}=−\mathrm{1}\:\wedge\:\mathrm{c}=\mathrm{2}\:\Rightarrow\:\mathrm{a}+\mathrm{b}−\mathrm{c}=\mathrm{0} \\ $$