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Question Number 41634 by Tawa1 last updated on 10/Aug/18
Commented by maxmathsup by imad last updated on 10/Aug/18
I = ∫ (dx/(sinx +(1/(cosx)))) = ∫ ((cosx)/(sinx .cosx +1))dx changement tan((x/2))=t give I = ∫ (((1−t^2 )/(1+t^2 ))/(((2t(1−t^2 ))/((1+t^2 )^2 )) +1)) ((2dt)/(1+t^2 )) = ∫ (((1−t^2 )/(1+t^2 ))/(((2t(1−t^2 ))/(1 +t^2 )) +1+t^2 )) 2dt = ∫ ((2(1−t^2 ))/(2t(1−t^2 ) +(1+t^2 )^2 ))dt = ∫ ((2−2t^2 )/(2t−2t^3 +t^4 +2t^2 +1)) dt = ∫ ((−2t^2 +2)/(t^4 −2t^3 +2t^2 +2t +1)) dt the roots of p(t)=t^4 −2t^3 +2t^2 +2t +1 are t_1 ∼ 1,366 +1,366 i =α(1+i) t_2 ∼ 1,366−1,366 i =α(1−i) with α =1,366 t_3 ∼−0,366 +0,366 i =β(1−i) and t_4 =β(1+i) with β =−0,366 t_4 =−0,366 −0,366i all roots are complex so F(t) = ((−2t^2 +2)/((t −α(1+i))(t−α(1−i))(t −β(1−i)(t−β(1+i))) = ((−2t^2 +2)/({(t−α)^2 +α^2 }{ (t−β)^2 +β^2 })) =((at +b)/((t−α)^2 +α^2 )) +((ct +d)/((t−β)^2 +β^2 )) ⇒ ∫ F(t)dt = ∫ ((at+b)/((t−α)^2 +α^2 )) dt + ∫ ((ct +d)/((t−β)^2 +β^2 )) dt +c... be continued...
$${I}\:\:=\:\int\:\:\:\:\frac{{dx}}{{sinx}\:+\frac{\mathrm{1}}{{cosx}}}\:=\:\:\int\:\:\:\:\:\:\frac{{cosx}}{{sinx}\:.{cosx}\:+\mathrm{1}}{dx}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}\:=\:\int\:\:\:\:\:\:\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:+\mathrm{1}}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\:=\:\:\int\:\:\:\:\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{1}\:+{t}^{\mathrm{2}} }\:+\mathrm{1}+{t}^{\mathrm{2}} }\:\mathrm{2}{dt} \\ $$$$=\:\:\int\:\:\:\:\:\:\:\:\frac{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\:+\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\:\int\:\:\:\:\frac{\mathrm{2}−\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} \:\:+{t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}}\:{dt} \\ $$$$=\:\int\:\:\:\:\:\frac{−\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}}{{t}^{\mathrm{4}} \:−\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:\:+\mathrm{2}{t}\:+\mathrm{1}}\:{dt}\:\:{the}\:{roots}\:{of}\:{p}\left({t}\right)={t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{3}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{1}\:{are} \\ $$$${t}_{\mathrm{1}} \sim\:\mathrm{1},\mathrm{366}\:+\mathrm{1},\mathrm{366}\:{i}\:=\alpha\left(\mathrm{1}+{i}\right)\:\: \\ $$$${t}_{\mathrm{2}} \sim\:\mathrm{1},\mathrm{366}−\mathrm{1},\mathrm{366}\:{i}\:=\alpha\left(\mathrm{1}−{i}\right)\:\:\:{with}\:\alpha\:=\mathrm{1},\mathrm{366} \\ $$$${t}_{\mathrm{3}} \:\sim−\mathrm{0},\mathrm{366}\:+\mathrm{0},\mathrm{366}\:{i}\:=\beta\left(\mathrm{1}−{i}\right)\:\:\:{and}\:{t}_{\mathrm{4}} =\beta\left(\mathrm{1}+{i}\right)\:{with}\:\beta\:=−\mathrm{0},\mathrm{366} \\ $$$${t}_{\mathrm{4}} =−\mathrm{0},\mathrm{366}\:−\mathrm{0},\mathrm{366}{i}\:\:{all}\:{roots}\:{are}\:{complex}\:\:{so} \\ $$$${F}\left({t}\right)\:=\:\:\frac{−\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}}{\left({t}\:−\alpha\left(\mathrm{1}+{i}\right)\right)\left({t}−\alpha\left(\mathrm{1}−{i}\right)\right)\left({t}\:−\beta\left(\mathrm{1}−{i}\right)\left({t}−\beta\left(\mathrm{1}+{i}\right)\right.\right.} \\ $$$$=\:\frac{−\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}}{\left\{\left({t}−\alpha\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} \right\}\left\{\:\left({t}−\beta\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right\}}\:=\frac{{at}\:+{b}}{\left({t}−\alpha\right)^{\mathrm{2}} \:+\alpha^{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{\left({t}−\beta\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int\:\:{F}\left({t}\right){dt}\:=\:\int\:\:\:\frac{{at}+{b}}{\left({t}−\alpha\right)^{\mathrm{2}} +\alpha^{\mathrm{2}} }\:{dt}\:+\:\int\:\:\:\frac{{ct}\:+{d}}{\left({t}−\beta\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:{dt}\:+{c}… \\ $$$${be}\:{continued}… \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Aug/18
∫((2cosx)/(2sinxcosx+2))dx ∫((cosx+sinx+cosx−sinx)/(1+1+2sinxcosx))dx ∫((cosx+sinx+cosx−sinx)/(1+cos^2 x+sin^2 x+2sinxcosx))dx ∫((cosx+sinx+cosx−sinx)/(1+(sinx+cosx)^2 ))dx ∫((cosx−sinx)/(1+(sinx+cosx)^2 ))dx+∫((−(cosx+sinx))/(−1−(sin^2 x+cos^2 x+2sinxcosx))dx ∫((d(sinx+cosx))/(1+(sinx+cosx)^2 ))−∫(((cosx+sinx)/(−1−1−2sinxcosx))dx ∫(dt/(1+t^2 ))−∫((cosx+sinx)/(−3+1−2sinxcosx))dx tan^(−1) (t)−∫((d(sinx−cosx))/(−3+(sinx−cosx)^2 )) tan^(−1) (sinx+cosx)−∫(dt_2 /(t_2 ^2 −((√3) )^2 )) ∫(dt_2 /((t_2 +(√3) )(t_2 −(√(3))))) =(1/(2(√3)))∫(((t_2 +(√3) )−(t_2 −(√3) ))/((t_2 +(√3) )(t−(√3) )))dt_2 (1/(2(√3))){∫(dt_2 /(t_2 −(√3) ))−∫(dt_2 /(t_2 +(√3) ))} (1/(2(√3) ))ln(((t_2 −(√3) )/(t_(2 ) +(√3))))+c ans is tan^(−1) (sinx+cosx)−(1/(2(√3) ))ln(((sinx−cosx−(√3))/(sinx−cosx+(√3))))+c
$$\int\frac{\mathrm{2}{cosx}}{\mathrm{2}{sinxcosx}+\mathrm{2}}{dx} \\ $$$$\int\frac{{cosx}+{sinx}+{cosx}−{sinx}}{\mathrm{1}+\mathrm{1}+\mathrm{2}{sinxcosx}}{dx} \\ $$$$\int\frac{{cosx}+{sinx}+{cosx}−{sinx}}{\mathrm{1}+{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}+\mathrm{2}{sinxcosx}}{dx} \\ $$$$\int\frac{{cosx}+{sinx}+{cosx}−{sinx}}{\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} }{dx} \\ $$$$\int\frac{{cosx}−{sinx}}{\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} }{dx}+\int\frac{−\left({cosx}+{sinx}\right)}{−\mathrm{1}−\left({sin}^{\mathrm{2}} {x}+{cos}^{\mathrm{2}} {x}+\mathrm{2}{sinxcosx}\right.}{dx} \\ $$$$\int\frac{{d}\left({sinx}+{cosx}\right)}{\mathrm{1}+\left({sinx}+{cosx}\right)^{\mathrm{2}} }−\int\frac{\left({cosx}+{sinx}\right.}{−\mathrm{1}−\mathrm{1}−\mathrm{2}{sinxcosx}}{dx} \\ $$$$\int\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\int\frac{{cosx}+{sinx}}{−\mathrm{3}+\mathrm{1}−\mathrm{2}{sinxcosx}}{dx} \\ $$$${tan}^{−\mathrm{1}} \left({t}\right)−\int\frac{{d}\left({sinx}−{cosx}\right)}{−\mathrm{3}+\left({sinx}−{cosx}\right)^{\mathrm{2}} } \\ $$$${tan}^{−\mathrm{1}} \left({sinx}+{cosx}\right)−\int\frac{{dt}_{\mathrm{2}} }{{t}_{\mathrm{2}} ^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\:\:\right)^{\mathrm{2}} } \\ $$$$\int\frac{{dt}_{\mathrm{2}} }{\left({t}_{\mathrm{2}} +\sqrt{\mathrm{3}}\:\right)\left({t}_{\mathrm{2}} −\sqrt{\left.\mathrm{3}\right)}\right.}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\int\frac{\left({t}_{\mathrm{2}} +\sqrt{\mathrm{3}}\:\right)−\left({t}_{\mathrm{2}} −\sqrt{\mathrm{3}}\:\right)}{\left({t}_{\mathrm{2}} +\sqrt{\mathrm{3}}\:\right)\left({t}−\sqrt{\mathrm{3}}\:\right)}{dt}_{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{\int\frac{{dt}_{\mathrm{2}} }{{t}_{\mathrm{2}} −\sqrt{\mathrm{3}}\:}−\int\frac{{dt}_{\mathrm{2}} }{{t}_{\mathrm{2}} +\sqrt{\mathrm{3}}\:}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\:}{ln}\left(\frac{{t}_{\mathrm{2}} \:−\sqrt{\mathrm{3}}\:}{{t}_{\mathrm{2}\:} +\sqrt{\mathrm{3}}}\right)+{c} \\ $$$${ans}\:{is} \\ $$$${tan}^{−\mathrm{1}} \left({sinx}+{cosx}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}\:}{ln}\left(\frac{{sinx}−{cosx}−\sqrt{\mathrm{3}}}{{sinx}−{cosx}+\sqrt{\mathrm{3}}}\right)+{c} \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa1 last updated on 10/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by Meritguide1234 last updated on 12/Aug/18
=∫((cosx)/(sinxcosx+1))dx =∫((2cosx)/(2+sin2x))dx =∫(((cosx+sinx)+(cosx−sinx))/(2+sin2x))dx =∫((d(sinx−cosx))/(3−(sinx−cosx)^2 ))+∫((d(cosx+sinx))/(1+(cosx+sinx)^2 ))
$$=\int\frac{{cosx}}{{sinxcosx}+\mathrm{1}}{dx} \\ $$$$=\int\frac{\mathrm{2}{cosx}}{\mathrm{2}+{sin}\mathrm{2}{x}}{dx} \\ $$$$=\int\frac{\left({cosx}+{sinx}\right)+\left({cosx}−{sinx}\right)}{\mathrm{2}+{sin}\mathrm{2}{x}}{dx} \\ $$$$=\int\frac{{d}\left({sinx}−{cosx}\right)}{\mathrm{3}−\left({sinx}−{cosx}\right)^{\mathrm{2}} }+\int\frac{{d}\left({cosx}+{sinx}\right)}{\mathrm{1}+\left({cosx}+{sinx}\right)^{\mathrm{2}} } \\ $$
Commented by Tawa1 last updated on 12/Aug/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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