Question Number 59631 by Mr X pcx last updated on 12/May/19
$$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{{acosx}\:+{bsinx}} \\ $$$${with}\:{a}\:,\:{b}\:{reals} \\ $$$$\left.\mathrm{2}\right){find}\:{also}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cosx}\:{dx}}{\left({acosx}\:+{bsinx}\right)^{\mathrm{2}} }\:\:{and} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sinx}\:{dx}}{\left({acosx}\:+{bsinx}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\mathrm{2}{cosx}\:+\sqrt{\mathrm{3}}{sinx}} \\ $$
Commented by Mr X pcx last updated on 12/May/19
$$\left.\mathrm{4}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cosx}\:{dx}}{\left(\mathrm{3}{cosx}\:+\sqrt{\mathrm{3}}{sinx}\right)^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 15/May/19
![we have I =∫_0 ^π (dx/(acosx +bsinx)) +∫_π ^(2π) (dx/(acosx +bsinx)) =H +K H =_(tan((x/2))=t) ∫_0 ^∞ (1/(a((1−t^2 )/(1+t^2 )) +b((2t)/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^∞ ((2dt)/(a−at^2 +2bt)) =∫_0 ^∞ ((−2dt)/(at^2 −2bt −a)) roots of at^2 −2bt −a =0 Δ^′ =b^2 +a^2 >0 for a ≠0 ⇒t_1 =((b+(√(a^2 +b^2 )))/a) t_2 =((b−(√(a^2 +b^2 )))/a) ⇒H =∫_0 ^∞ ((−2)/(a(t−t_1 )(t−t_2 ))) dt =((−2)/(a(t_1 −t_2 )))∫_0 ^∞ ((1/(t−t_1 )) −(1/(t−t_2 )))dt =((−2)/((2a(√(a^2 +b^2 )))/a)) [ln∣((t−t_1 )/(t−t_2 ))∣]_0 ^(+∞) =((−1)/( (√(a^2 +b^2 ))))(−ln∣(t_1 /t_2 )∣) =(1/( (√(a^2 +b^2 )))) ln∣((b+(√(a^2 +b^2 )))/(b−(√(a^2 +b^2 ))))∣ ⇒H =(1/( (√(a^2 +b^2 ))))ln((((√(a^2 +b^2 )) +b)/( (√(a^2 +b^2 )) −b))) K =_(x =t +π) ∫_0 ^π (dt/(−a cost −bsint)) =−(1/( (√(a^2 +b^2 )))) ln((((√(a^2 +b^2 )) +b)/( (√(a^2 +b^2 ))−b))) ⇒ I =0](https://www.tinkutara.com/question/Q59878.png)
$${we}\:{have}\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{{acosx}\:+{bsinx}}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{{acosx}\:+{bsinx}}\:={H}\:+{K} \\ $$$${H}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{{a}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+{b}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{{a}−{at}^{\mathrm{2}} \:+\mathrm{2}{bt}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{−\mathrm{2}{dt}}{{at}^{\mathrm{2}} \:−\mathrm{2}{bt}\:−{a}}\:\:\:{roots}\:{of}\:\:{at}^{\mathrm{2}} −\mathrm{2}{bt}\:−{a}\:=\mathrm{0} \\ $$$$\Delta^{'} \:={b}^{\mathrm{2}} \:+{a}^{\mathrm{2}} >\mathrm{0}\:\:\:{for}\:{a}\:\neq\mathrm{0}\:\:\Rightarrow{t}_{\mathrm{1}} =\frac{{b}+\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}} \\ $$$${t}_{\mathrm{2}} =\frac{{b}−\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}}\:\Rightarrow{H}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{−\mathrm{2}}{{a}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}\:{dt}\:=\frac{−\mathrm{2}}{{a}\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)}\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{1}}{{t}−{t}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{t}−{t}_{\mathrm{2}} }\right){dt} \\ $$$$=\frac{−\mathrm{2}}{\frac{\mathrm{2}{a}\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{{a}}}\:\left[{ln}\mid\frac{{t}−{t}_{\mathrm{1}} }{{t}−{t}_{\mathrm{2}} }\mid\right]_{\mathrm{0}} ^{+\infty} \:=\frac{−\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\left(−{ln}\mid\frac{{t}_{\mathrm{1}} }{{t}_{\mathrm{2}} }\mid\right)\:=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\:{ln}\mid\frac{{b}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{b}−\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\mid \\ $$$$\Rightarrow{H}\:=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}{ln}\left(\frac{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:+{b}}{\:\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:−{b}}\right) \\ $$$${K}\:=_{{x}\:={t}\:+\pi} \:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dt}}{−{a}\:{cost}\:−{bsint}}\:=−\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }}\:{ln}\left(\frac{\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }\:+{b}}{\:\sqrt{{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} }−{b}}\right)\:\Rightarrow \\ $$$${I}\:=\mathrm{0} \\ $$
Commented by maxmathsup by imad last updated on 15/May/19
$$\left.\mathrm{2}\right)\:{let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{{acosx}\:+{bsinx}}\:\Rightarrow{f}^{'} \left({a}\right)\:=−\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cosx}}{\left({acosx}\:+{bsinx}\right)^{\mathrm{2}} }{dx} \\ $$$${but}\:{f}\left({a}\right)=\mathrm{0}\:\Rightarrow{f}^{'} \left({a}\right)=\mathrm{0}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cosx}}{\left({acosx}\:+{bsinx}\right)^{\mathrm{2}} }{dx}\:=\mathrm{0}\:{also}\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sinx}}{\left({acosx}\:+{bsinx}\right)^{\mathrm{2}} }{dx}=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\mathrm{2}{cosx}\:+\sqrt{\mathrm{3}}{sinx}}\:=\mathrm{0}\:. \\ $$
Answered by tanmay last updated on 12/May/19
$$\int\frac{{dx}}{{acosx}+{bsinx}} \\ $$$$\int\frac{{dx}}{{rsin}\alpha{cosx}+{rcos}\alpha{sinx}} \\ $$$$\frac{\mathrm{1}}{{r}}\int\frac{{dx}}{{sin}\left(\alpha+{x}\right)} \\ $$$$\frac{\mathrm{1}}{{r}}\int{cosec}\left({x}+\alpha\right){dx} \\ $$$$\frac{\mathrm{1}}{{r}}×{lntan}\left(\frac{{x}+\alpha}{\mathrm{2}}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:}{lntan}\left(\frac{{x}+{tan}^{−\mathrm{1}} \left(\frac{{a}}{{b}}\right)}{\mathrm{2}}\right)+{c} \\ $$$${now}\:{proceed} \\ $$
Answered by tanmay last updated on 13/May/19
![2)∫((cosx)/((acosx+bsinx)^2 )) cosx=A(acosx+bsinx)+B×(d/dx)(acosx+bsinx) cosx=A(acosx+bsinx)+B×(−asinx+bcosx) cosx=(Aa+bB)cosx+(Ab−Ba)sinx Ab−Ba=0 (A/a)=(B/b)=k →A=ak and B=bk Aa+Bb=1 a^2 k+b^2 k=1 k=(1/(a^2 +b^2 )) so A=(a/(a^2 +b^2 )) and B=(b/(a^2 +b^2 )) now ∫((cosxdx)/((acosx+bsinx)^2 )) ∫[((A(acosx+bsinx)+B×(d/dx)(((acosx+bsinx))/))/((acosx+bsinx)^2 ))]dx ∫(A/((acosx+bsinx)))dx+B∫((d(acosx+bsinx))/((acosx+bsinx)^2 )) =(a/(a^2 +b^2 ))×(1/( (√(a^2 +b^2 ))))lntan(((x+tan^(−1) ((a/b)))/2))+(b/(a^2 +b^2 ))×((−1)/((acosx+bsinx)))+c now pls put the upper and lower limit for ∫((sinxdx)/((acosx+bsinx)^2 ))→same method...](https://www.tinkutara.com/question/Q59654.png)
$$\left.\mathrm{2}\right)\int\frac{{cosx}}{\left({acosx}+{bsinx}\right)^{\mathrm{2}} } \\ $$$${cosx}={A}\left({acosx}+{bsinx}\right)+{B}×\frac{{d}}{{dx}}\left({acosx}+{bsinx}\right) \\ $$$${cosx}={A}\left({acosx}+{bsinx}\right)+{B}×\left(−{asinx}+{bcosx}\right) \\ $$$${cosx}=\left({Aa}+{bB}\right){cosx}+\left({Ab}−{Ba}\right){sinx} \\ $$$${Ab}−{Ba}=\mathrm{0} \\ $$$$\frac{{A}}{{a}}=\frac{{B}}{{b}}={k}\:\rightarrow{A}={ak}\:\:\:{and}\:{B}={bk} \\ $$$${Aa}+{Bb}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} {k}+{b}^{\mathrm{2}} {k}=\mathrm{1} \\ $$$${k}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\:{so}\:{A}=\frac{{a}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:{and}\:{B}=\frac{{b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${now} \\ $$$$\int\frac{{cosxdx}}{\left({acosx}+{bsinx}\right)^{\mathrm{2}} } \\ $$$$\int\left[\frac{{A}\left({acosx}+{bsinx}\right)+{B}×\frac{{d}}{{dx}}\frac{\left({acosx}+{bsinx}\right)}{}}{\left({acosx}+{bsinx}\right)^{\mathrm{2}} }\right]{dx} \\ $$$$\int\frac{{A}}{\left({acosx}+{bsinx}\right)}{dx}+{B}\int\frac{{d}\left({acosx}+{bsinx}\right)}{\left({acosx}+{bsinx}\right)^{\mathrm{2}} } \\ $$$$=\frac{{a}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }×\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{lntan}\left(\frac{{x}+{tan}^{−\mathrm{1}} \left(\frac{{a}}{{b}}\right)}{\mathrm{2}}\right)+\frac{{b}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }×\frac{−\mathrm{1}}{\left({acosx}+{bsinx}\right)}+{c} \\ $$$${now}\:{pls}\:{put}\:{the}\:{upper}\:{and}\:{lower}\:{limit} \\ $$$${for}\:\int\frac{{sinxdx}}{\left({acosx}+{bsinx}\right)^{\mathrm{2}} }\rightarrow{same}\:{method}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$