Question Number 27757 by abdo imad last updated on 14/Jan/18
$${calculate}\:\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+{cosx}}\:{and}\:{J}=\:\int_{\mathrm{0}^{} } ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{cosx}}{\mathrm{1}+{cosx}}{dx}\:. \\ $$
Answered by mrW2 last updated on 14/Jan/18
![∫(dx/(1+cos x))=∫(1/(cos^2 (x/2))) d((x/2))=tan (x/2)+C ∫((cos x dx)/(1+cos x))=∫((1+cos x−1)/(1+cos x)) dx=x−∫(dx/(1+cos x))=x−tan (x/2)+C ⇒I=[tan (x/2)]_0 ^(π/2) =tan (π/4)=1 ⇒J=[x−tan (x/2)]_0 ^(π/2) =(π/2)−tan (π/4)=(π/2)−1](https://www.tinkutara.com/question/Q27763.png)
$$\int\frac{{dx}}{\mathrm{1}+\mathrm{cos}\:{x}}=\int\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}\:{d}\left(\frac{{x}}{\mathrm{2}}\right)=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}+{C} \\ $$$$\int\frac{\mathrm{cos}\:{x}\:{dx}}{\mathrm{1}+\mathrm{cos}\:{x}}=\int\frac{\mathrm{1}+\mathrm{cos}\:{x}−\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:{x}}\:{dx}={x}−\int\frac{{dx}}{\mathrm{1}+\mathrm{cos}\:{x}}={x}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}+{C} \\ $$$$\Rightarrow{I}=\left[\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\mathrm{tan}\:\frac{\pi}{\mathrm{4}}=\mathrm{1} \\ $$$$\Rightarrow{J}=\left[{x}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}}−\mathrm{tan}\:\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}}−\mathrm{1} \\ $$