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1-1-cos-x-dx-2-1-2-cot-x-dx-3-1-ln-cos-x-dx-

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Question Number 102690 by bramlex last updated on 10/Jul/20
(1)∫(1/(cos (√x))) dx (2) ∫ (1/(2+cot x)) dx (3) ∫ (1/(ln(cos x))) dx
$$\left(\mathrm{1}\right)\int\frac{\mathrm{1}}{\mathrm{cos}\:\sqrt{\mathrm{x}}}\:\mathrm{dx}\: \\ $$$$\left(\mathrm{2}\right)\:\int\:\frac{\mathrm{1}}{\mathrm{2}+\mathrm{cot}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$$$\left(\mathrm{3}\right)\:\int\:\frac{\mathrm{1}}{\mathrm{ln}\left(\mathrm{cos}\:\mathrm{x}\right)}\:\mathrm{dx}\: \\ $$
Answered by Dwaipayan Shikari last updated on 10/Jul/20
2)∫((tanx)/(2tanx+1))=(1/2)∫((2tanx+1)/(2tanx+1))−(1/2)∫(1/(2tanx+1))=(x/2)−(1/2)∫(2/((2t+1)(t^2 +1)))dt =(x/2)−∫((At+C)/(t^2 +1))+(B/(2t+1)) {(1/((2t+1)(t^2 +1)))=((At+C)/(t^2 +1))+(B/(2t+1)) and take (t=tanx) =(x/2)−I_a {2At^2 +At+2Ct+C+Bt^2 +B=1 =(x/2)+∫(((2/5)t−(1/5))/(t^2 +1))−((4/5)/(2t+1)) { 2A+B=0 A+2C=0 B+C=1 =(x/2)+(1/5)∫((2t−1)/(t^2 +1))−(4/(10))∫(1/(t+(1/2)))dt {A=−(2/5) B=(4/5) C=(1/5) =(x/2)+(1/5)∫((2t)/(t^2 +1))−(1/5)tan^(−1) tanx−(4/(10))log(t+(1/2)) =(x/2)+(1/5)log(t^2 +1)−(1/5)x−(4/(10))log(tanx+(1/2)) =(x/2)+(2/5)log(secx)−(1/5)x−(4/(10))log(tanx+(1/2))+Constant =(1/5)(2x−log(cosx)−2log(tanx+(1/2)))+Constant
$$\left.\mathrm{2}\right)\int\frac{{tanx}}{\mathrm{2}{tanx}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{tanx}+\mathrm{1}}{\mathrm{2}{tanx}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{2}{tanx}+\mathrm{1}}=\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}}{\left(\mathrm{2}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\ $$$$ \\ $$$$=\frac{{x}}{\mathrm{2}}−\int\frac{{At}+{C}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{B}}{\mathrm{2}{t}+\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\left\{\frac{\mathrm{1}}{\left(\mathrm{2}{t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{At}+{C}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{B}}{\mathrm{2}{t}+\mathrm{1}}\:\:{and}\:{take}\right. \\ $$$$\left({t}={tanx}\right) \\ $$$$=\frac{{x}}{\mathrm{2}}−{I}_{{a}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\mathrm{2}{At}^{\mathrm{2}} +{At}+\mathrm{2}{Ct}+{C}+{Bt}^{\mathrm{2}} +{B}=\mathrm{1}\right. \\ $$$$=\frac{{x}}{\mathrm{2}}+\int\frac{\frac{\mathrm{2}}{\mathrm{5}}{t}−\frac{\mathrm{1}}{\mathrm{5}}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\frac{\mathrm{4}}{\mathrm{5}}}{\mathrm{2}{t}+\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\:\:\mathrm{2}{A}+{B}=\mathrm{0}\:\:\:\:{A}+\mathrm{2}{C}=\mathrm{0}\:\:{B}+{C}=\mathrm{1}\right. \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{2}{t}−\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{4}}{\mathrm{10}}\int\frac{\mathrm{1}}{{t}+\frac{\mathrm{1}}{\mathrm{2}}}{dt}\:\:\:\:\:\left\{{A}=−\frac{\mathrm{2}}{\mathrm{5}}\:\:{B}=\frac{\mathrm{4}}{\mathrm{5}}\:\:{C}=\frac{\mathrm{1}}{\mathrm{5}}\right. \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}}\int\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{5}}{tan}^{−\mathrm{1}} {tanx}−\frac{\mathrm{4}}{\mathrm{10}}{log}\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{5}}{log}\left({t}^{\mathrm{2}} +\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{5}}{x}−\frac{\mathrm{4}}{\mathrm{10}}{log}\left({tanx}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{5}}{log}\left({secx}\right)−\frac{\mathrm{1}}{\mathrm{5}}{x}−\frac{\mathrm{4}}{\mathrm{10}}{log}\left({tanx}+\frac{\mathrm{1}}{\mathrm{2}}\right)+{Constant} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{2}{x}−{log}\left({cosx}\right)−\mathrm{2}{log}\left({tanx}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)+{Constant} \\ $$
Answered by mathmax by abdo last updated on 10/Jul/20
2) I =∫ (dx/(2+((cosx)/(sinx)))) ⇒ I =∫ ((sinx)/(2sinx +cosx))dx changement tan((x/2))=t give I =∫ (((2t)/(1+t^2 ))/(2×((2t)/(1+t^2 )) +((1−t^2 )/(1+t^2 )))) dt =∫ ((2t)/(4t+1−t^2 )) dt =−2 ∫ ((tdt)/(t^2 −4t−1)) t^2 −4t−1 =0 →Δ^′ =4+1 =5 ⇒t_1 =2+(√5) and t_2 =2−(√5) ⇒ F(t) =(t/(t^2 −4t−1)) =(t/((t−t_1 )(t−t_2 ))) =(a/(t−t_1 )) +(b/(t−t_2 )) a =(t_1 /(t_1 −t_2 )) =((2+(√5))/(2(√5))) , b =(t_2 /(t_2 −t_1 )) =((2−(√5))/(−2(√5))) =((−2+(√5))/(2(√5))) ⇒ I =−2 ∫ ((adt)/(t−t_1 )) −2 ∫ ((bdt)/(t−t_2 )) =−2aln∣t−t_1 ∣ −2 b ln∣t−t_2 ∣ +c =−2×((2+(√5))/(2%5))ln∣tan((x/2))−2−(√5)∣−2×((−2+(√5))/(2(√5)))ln∣tan((x/2))−2+(√5)∣ +C =(((−2−(√5))/( (√5))))ln∣tan((x/2))−2−(√5)∣ +((2−(√5))/( (√5)))ln∣tan((x/2))−2+(√5)∣ +C
$$\left.\mathrm{2}\right)\:\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\mathrm{2}+\frac{\mathrm{cosx}}{\mathrm{sinx}}}\:\Rightarrow\:\mathrm{I}\:=\int\:\:\frac{\mathrm{sinx}}{\mathrm{2sinx}\:+\mathrm{cosx}}\mathrm{dx}\:\mathrm{changement}\:\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}\:=\int\:\:\frac{\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}{\mathrm{2}×\frac{\mathrm{2t}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }}\:\mathrm{dt}\:=\int\:\:\frac{\mathrm{2t}}{\mathrm{4t}+\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:\mathrm{dt}\:=−\mathrm{2}\:\int\:\:\frac{\mathrm{tdt}}{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}−\mathrm{1}} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{4t}−\mathrm{1}\:=\mathrm{0}\:\rightarrow\Delta^{'} \:=\mathrm{4}+\mathrm{1}\:=\mathrm{5}\:\Rightarrow\mathrm{t}_{\mathrm{1}} =\mathrm{2}+\sqrt{\mathrm{5}}\:\mathrm{and}\:\mathrm{t}_{\mathrm{2}} =\mathrm{2}−\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\mathrm{t}}{\mathrm{t}^{\mathrm{2}} −\mathrm{4t}−\mathrm{1}}\:=\frac{\mathrm{t}}{\left(\mathrm{t}−\mathrm{t}_{\mathrm{1}} \right)\left(\mathrm{t}−\mathrm{t}_{\mathrm{2}} \right)}\:=\frac{\mathrm{a}}{\mathrm{t}−\mathrm{t}_{\mathrm{1}} }\:+\frac{\mathrm{b}}{\mathrm{t}−\mathrm{t}_{\mathrm{2}} } \\ $$$$\mathrm{a}\:=\frac{\mathrm{t}_{\mathrm{1}} }{\mathrm{t}_{\mathrm{1}} −\mathrm{t}_{\mathrm{2}} }\:=\frac{\mathrm{2}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\:,\:\mathrm{b}\:=\frac{\mathrm{t}_{\mathrm{2}} }{\mathrm{t}_{\mathrm{2}} −\mathrm{t}_{\mathrm{1}} }\:=\frac{\mathrm{2}−\sqrt{\mathrm{5}}}{−\mathrm{2}\sqrt{\mathrm{5}}}\:=\frac{−\mathrm{2}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$$\mathrm{I}\:=−\mathrm{2}\:\int\:\:\:\frac{\mathrm{adt}}{\mathrm{t}−\mathrm{t}_{\mathrm{1}} }\:−\mathrm{2}\:\int\:\frac{\mathrm{bdt}}{\mathrm{t}−\mathrm{t}_{\mathrm{2}} } \\ $$$$=−\mathrm{2aln}\mid\mathrm{t}−\mathrm{t}_{\mathrm{1}} \mid\:−\mathrm{2}\:\mathrm{b}\:\mathrm{ln}\mid\mathrm{t}−\mathrm{t}_{\mathrm{2}} \mid\:+\mathrm{c} \\ $$$$=−\mathrm{2}×\frac{\mathrm{2}+\sqrt{\mathrm{5}}}{\mathrm{2\%5}}\mathrm{ln}\mid\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{2}−\sqrt{\mathrm{5}}\mid−\mathrm{2}×\frac{−\mathrm{2}+\sqrt{\mathrm{5}}}{\mathrm{2}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{2}+\sqrt{\mathrm{5}}\mid\:+\mathrm{C} \\ $$$$=\left(\frac{−\mathrm{2}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}}\right)\mathrm{ln}\mid\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{2}−\sqrt{\mathrm{5}}\mid\:+\frac{\mathrm{2}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}}\mathrm{ln}\mid\mathrm{tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{2}+\sqrt{\mathrm{5}}\mid\:+\mathrm{C} \\ $$
Answered by bemath last updated on 11/Jul/20

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