Question Number 32351 by abdo imad last updated on 23/Mar/18
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{{dt}}{\mathrm{1}+{cos}\theta\:{sint}}\:.\: \\ $$
Answered by sma3l2996 last updated on 25/Mar/18
![let x=tan(t/2)⇒dt=((2dx)/(1+x^2 )) sint=((2x)/(1+x^2 )) ∫_0 ^(π/2) (dt/(1+cosθsint))=2∫_0 ^1 (dx/(1+x^2 +2xcosθ))=2∫_0 ^1 (dx/((x+cosθ)^2 +sin^2 θ)) =(2/(sin^2 θ))∫_0 ^1 (dx/((((x+cosθ)/(sinθ)))^2 +1)) let u=((x+cosθ)/(sinθ))⇒dx=sinθdu ∫_0 ^(π/2) (dt/(1+cosθsint))=(2/(sinθ))∫_(cotθ) ^((1+cosθ)/(sinθ)) (du/(u^2 +1))=(2/(sinθ))[arctan(((x+cosθ)/(sinθ)))]_0 ^1 =(2/(sinθ))(arctan(((1+cosθ)/(sinθ)))−arctan(cotθ))](https://www.tinkutara.com/question/Q32468.png)
$${let}\:{x}={tan}\left({t}/\mathrm{2}\right)\Rightarrow{dt}=\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${sint}=\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dt}}{\mathrm{1}+{cos}\theta{sint}}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} +\mathrm{2}{xcos}\theta}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left({x}+{cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta} \\ $$$$=\frac{\mathrm{2}}{{sin}^{\mathrm{2}} \theta}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\frac{{x}+{cos}\theta}{{sin}\theta}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${let}\:{u}=\frac{{x}+{cos}\theta}{{sin}\theta}\Rightarrow{dx}={sin}\theta{du} \\ $$$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{dt}}{\mathrm{1}+{cos}\theta{sint}}=\frac{\mathrm{2}}{{sin}\theta}\int_{{cot}\theta} ^{\frac{\mathrm{1}+{cos}\theta}{{sin}\theta}} \frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}}{{sin}\theta}\left[{arctan}\left(\frac{{x}+{cos}\theta}{{sin}\theta}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{{sin}\theta}\left({arctan}\left(\frac{\mathrm{1}+{cos}\theta}{{sin}\theta}\right)−{arctan}\left({cot}\theta\right)\right) \\ $$$$ \\ $$