Question Number 64139 by Chi Mes Try last updated on 14/Jul/19
$$\underset{\:\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}\:\mathrm{cos}\:\alpha+\mathrm{1}}\:=\:\alpha\:\mathrm{sin}\:\alpha \\ $$
Commented by Prithwish sen last updated on 14/Jul/19
![∫_0 ^1 (dx/((x+cosα)^2 +(sinα)^2 )) =[(1/((sinα))).tan^(−1) (((x+cos)/(sinα)))]_0 ^1](https://www.tinkutara.com/question/Q64142.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{cos}\alpha\right)^{\mathrm{2}} +\left(\mathrm{sin}\alpha\right)^{\mathrm{2}} } \\ $$$$=\left[\frac{\mathrm{1}}{\left(\mathrm{sin}\alpha\right)}.\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}+\mathrm{cos}}{\mathrm{sin}\alpha}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$
Commented by mathmax by abdo last updated on 14/Jul/19
![let I =∫_0 ^1 (dx/(x^2 +2xcosα +1)) ⇒I =∫_0 ^1 (dx/(x^2 +2xcosα +cos^2 α +sin^2 α)) =∫_0 ^1 (dx/((x+cosα)^2 +sin^2 α)) changement x+cosα =sinα t give I = ∫_(cotanα) ^((1+cosα)/(sinα)) ((sinα dt)/(sin^2 α(1+t^2 ))) =(1/(sinα))[arctant]_(1/(tanα)) ^((1+cosα)/(sinα)) =(1/(sinα)){ arctan(((1+cosα)/(sinα)))−arctan((1/(tanα)))but ((1+cosα)/(sinα)) =((2cos^2 ((α/2)))/(2sin((α/2))cos((α/2)))) =(1/(tan((α/2)))) ⇒ arctan(((1+cosα)/(sinα)))=+^− (π/2) −(α/2) arctan((1/(tanα)))=+^− (π/2) −α ⇒ I =(1/(sinα)){+^− (π/2) +^− (π/2) +(α/2)} so there is a error in the question if not a equation...](https://www.tinkutara.com/question/Q64145.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{xcos}\alpha\:+\mathrm{1}}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{xcos}\alpha\:+{cos}^{\mathrm{2}} \alpha\:+{sin}^{\mathrm{2}} \alpha} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left({x}+{cos}\alpha\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \alpha}\:\:{changement}\:{x}+{cos}\alpha\:={sin}\alpha\:{t}\:{give} \\ $$$${I}\:=\:\int_{{cotan}\alpha} ^{\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}} \:\:\:\frac{{sin}\alpha\:{dt}}{{sin}^{\mathrm{2}} \alpha\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{sin}\alpha}\left[{arctant}\right]_{\frac{\mathrm{1}}{{tan}\alpha}} ^{\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}} \\ $$$$=\frac{\mathrm{1}}{{sin}\alpha}\left\{\:{arctan}\left(\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}\right)−{arctan}\left(\frac{\mathrm{1}}{{tan}\alpha}\right){but}\right. \\ $$$$\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}\:=\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right){cos}\left(\frac{\alpha}{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{{tan}\left(\frac{\alpha}{\mathrm{2}}\right)}\:\Rightarrow \\ $$$${arctan}\left(\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}\right)=\overset{−} {+}\frac{\pi}{\mathrm{2}}\:−\frac{\alpha}{\mathrm{2}} \\ $$$${arctan}\left(\frac{\mathrm{1}}{{tan}\alpha}\right)=\overset{−} {+}\frac{\pi}{\mathrm{2}}\:−\alpha\:\:\Rightarrow\:{I}\:=\frac{\mathrm{1}}{{sin}\alpha}\left\{\overset{−} {+}\frac{\pi}{\mathrm{2}}\:\overset{−} {+}\frac{\pi}{\mathrm{2}}\:+\frac{\alpha}{\mathrm{2}}\right\} \\ $$$${so}\:{there}\:{is}\:{a}\:{error}\:{in}\:{the}\:{question}\:{if}\:{not}\:{a}\:{equation}… \\ $$
Answered by Hope last updated on 15/Jul/19
![D_r =(x+cosα)^2 +sin^2 α t=x+cosα ∫_(cosα) ^(1+cosα) (dt/(t^2 +sin^2 α)) (1/(sinα))∣tan^(−1) ((t/(sinα)))∣_(cosα) ^(1+cosα) =(1/(sinα))[tan^(−1) (((1+cosα)/(sinα)))−tan^(−1) (((cosα)/(sinα)))] =(1/(sinα))[tan^(−1) (cot(α/2))−tan^(−1) (cotα)] =(1/(sinα))[tan^(−1) (tan((π/2)−(α/2)))−tan^(−1) ((π/2)−α)] =(1/(sinα))[(π/2)−(α/2)−(π/2)+α] =(α/(2sinα))](https://www.tinkutara.com/question/Q64190.png)
$${D}_{{r}} =\left({x}+{cos}\alpha\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \alpha \\ $$$${t}={x}+{cos}\alpha \\ $$$$\int_{{cos}\alpha} ^{\mathrm{1}+{cos}\alpha} \frac{{dt}}{{t}^{\mathrm{2}} +{sin}^{\mathrm{2}} \alpha} \\ $$$$\frac{\mathrm{1}}{{sin}\alpha}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}}{{sin}\alpha}\right)\mid_{{cos}\alpha} ^{\mathrm{1}+{cos}\alpha} \\ $$$$=\frac{\mathrm{1}}{{sin}\alpha}\left[{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}\right)−{tan}^{−\mathrm{1}} \left(\frac{{cos}\alpha}{{sin}\alpha}\right)\right] \\ $$$$=\frac{\mathrm{1}}{{sin}\alpha}\left[{tan}^{−\mathrm{1}} \left({cot}\frac{\alpha}{\mathrm{2}}\right)−{tan}^{−\mathrm{1}} \left({cot}\alpha\right)\right] \\ $$$$=\frac{\mathrm{1}}{{sin}\alpha}\left[{tan}^{−\mathrm{1}} \left({tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}\right)\right)−{tan}^{−\mathrm{1}} \left(\frac{\pi}{\mathrm{2}}−\alpha\right)\right] \\ $$$$=\frac{\mathrm{1}}{{sin}\alpha}\left[\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}+\alpha\right] \\ $$$$=\frac{\alpha}{\mathrm{2}{sin}\alpha} \\ $$