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Question Number 46238 by Meritguide1234 last updated on 23/Oct/18

Commented by math1967 last updated on 23/Oct/18

cotxcot2xcot3x=cotx+cot2x+cot3x

$${cotxcot}\mathrm{2}{xcot}\mathrm{3}{x}={cotx}+{cot}\mathrm{2}{x}+{cot}\mathrm{3}{x} \\ $$

Commented by Meritguide1234 last updated on 23/Oct/18

no

$${no} \\ $$

Commented by math1967 last updated on 23/Oct/18

cot3x=((cot2xcotx−1)/(cot2x+cotx)) ∴cot3xcot2x +cot3xcotx=cot2xcotx−1 so that was error my comment

$${cot}\mathrm{3}{x}=\frac{{cot}\mathrm{2}{xcotx}−\mathrm{1}}{{cot}\mathrm{2}{x}+{cotx}} \\ $$$$\therefore{cot}\mathrm{3}{xcot}\mathrm{2}{x}\:+{cot}\mathrm{3}{xcotx}={cot}\mathrm{2}{xcotx}−\mathrm{1} \\ $$$${so}\:{that}\:{was}\:{error}\:{my}\:{comment} \\ $$

Commented by MJS last updated on 23/Oct/18

why not try these: 5. ∫sec x sec 2x sec 3x dx 6. ∫csc x csc 2x csc 3x dx

$$\mathrm{why}\:\mathrm{not}\:\mathrm{try}\:\mathrm{these}: \\ $$$$\mathrm{5}.\:\int\mathrm{sec}\:{x}\:\mathrm{sec}\:\mathrm{2}{x}\:\mathrm{sec}\:\mathrm{3}{x}\:{dx} \\ $$$$\mathrm{6}.\:\int\mathrm{csc}\:{x}\:\mathrm{csc}\:\mathrm{2}{x}\:\mathrm{csc}\:\mathrm{3}{x}\:{dx} \\ $$

Commented by Meritguide1234 last updated on 23/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

3)tan3x=tan(x+2x) =((tanx+tan2x)/(1−tanx.tan2x)) tan3x−tanxtan2xtan3x=tanx+tan2x tan3x−tanx−tan2x=tanxtan2xtan3x so∫tanxtan2xtan3xdx =∫(tan3x−tan2x−tanx) dx =(1/3)×lnsec3x−(1/2)×lnsec2x−lnsecx +c =corrected

$$\left.\mathrm{3}\right){tan}\mathrm{3}{x}={tan}\left({x}+\mathrm{2}{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{tanx}+{tan}\mathrm{2}{x}}{\mathrm{1}−{tanx}.{tan}\mathrm{2}{x}} \\ $$$${tan}\mathrm{3}{x}−{tanxtan}\mathrm{2}{xtan}\mathrm{3}{x}={tanx}+{tan}\mathrm{2}{x} \\ $$$${tan}\mathrm{3}{x}−{tanx}−{tan}\mathrm{2}{x}={tanxtan}\mathrm{2}{xtan}\mathrm{3}{x} \\ $$$${so}\int{tanxtan}\mathrm{2}{xtan}\mathrm{3}{xdx} \\ $$$$=\int\left({tan}\mathrm{3}{x}−{tan}\mathrm{2}{x}−{tanx}\right)\:\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}×{lnsec}\mathrm{3}{x}−\frac{\mathrm{1}}{\mathrm{2}}×{lnsec}\mathrm{2}{x}−{lnsecx}\:\:+{c} \\ $$$$={corrected} \\ $$

Commented by Meritguide1234 last updated on 23/Oct/18

last line integrand is (1/3)lnsec3x−(1/2)lnsec2x−lnsecx+c

$${last}\:{line}\:{integrand}\:{is}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}{lnsec}\mathrm{3}{x}−\frac{\mathrm{1}}{\mathrm{2}}{lnsec}\mathrm{2}{x}−{lnsecx}+{c} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

yes in hurry...

$${yes}\:{in}\:{hurry}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

1)∫sinxsin2xsin3x dx =(1/2)∫sinx(2sin2xsin3x)dx =(1/2)∫sinx(cosx−cos5x)dx =(1/4)∫{(2sinxcosx)−(2sinxcos5x)}dx =(1/4)∫{(sin2x−(sin6x−sin4x)}dx =(1/4)∫(sin2x−sin6x+sin4x)dx =(1/4){((−cos2x)/2)+((cos6x)/6)+((−cos4x)/4)}+c

$$\left.\mathrm{1}\right)\int{sinxsin}\mathrm{2}{xsin}\mathrm{3}{x}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{sinx}\left(\mathrm{2}{sin}\mathrm{2}{xsin}\mathrm{3}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{sinx}\left({cosx}−{cos}\mathrm{5}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left\{\left(\mathrm{2}{sinxcosx}\right)−\left(\mathrm{2}{sinxcos}\mathrm{5}{x}\right)\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left\{\left({sin}\mathrm{2}{x}−\left({sin}\mathrm{6}{x}−{sin}\mathrm{4}{x}\right)\right\}{dx}\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left({sin}\mathrm{2}{x}−{sin}\mathrm{6}{x}+{sin}\mathrm{4}{x}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{−{cos}\mathrm{2}{x}}{\mathrm{2}}+\frac{{cos}\mathrm{6}{x}}{\mathrm{6}}+\frac{−{cos}\mathrm{4}{x}}{\mathrm{4}}\right\}+{c} \\ $$$$ \\ $$$$ \\ $$

Commented by Meritguide1234 last updated on 23/Oct/18

correct

$${correct} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

2)(1/2)∫cosx(2cos2xcos3x)dx =(1/2)∫cosx(cos5x+cosx)dx =(1/4)∫{2cosxcos5x+2cos^2 x}dx =(1/4)∫{cos6x+cos4x+cos2x+1}dx (1/4){((sin6x)/6)+((sin4x)/4)+((sin2x)/2)+x}+c

$$\left.\mathrm{2}\right)\frac{\mathrm{1}}{\mathrm{2}}\int{cosx}\left(\mathrm{2}{cos}\mathrm{2}{xcos}\mathrm{3}{x}\right){dx} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int{cosx}\left({cos}\mathrm{5}{x}+{cosx}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left\{\mathrm{2}{cosxcos}\mathrm{5}{x}+\mathrm{2}{cos}^{\mathrm{2}} {x}\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left\{{cos}\mathrm{6}{x}+{cos}\mathrm{4}{x}+{cos}\mathrm{2}{x}+\mathrm{1}\right\}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{{sin}\mathrm{6}{x}}{\mathrm{6}}+\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}+\frac{{sin}\mathrm{2}{x}}{\mathrm{2}}+{x}\right\}+{c} \\ $$

Commented by Meritguide1234 last updated on 23/Oct/18

correct...try no 4

$${correct}...{try}\:{no}\:\mathrm{4} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

yes sir i am trying Q 4...

$${yes}\:{sir}\:{i}\:{am}\:{trying}\:{Q}\:\mathrm{4}... \\ $$

Answered by MJS last updated on 23/Oct/18

∫cot x cot 2x cot 3x dx= =∫(cot x ×((cos^2 x −sin^2 x)/(2cos x sin x))×((cos x (cos^2 x −3sin^2 x))/(sin x (3cos^2 x −sin^2 x)))dx= [cos x =((cot x)/(csc x)); sin x =(1/(csc x)); csc^2 x =1+cot^2 x] =∫((csc^2 x cot x (cot^4 x −4cot^2 x +3))/(2(3cot^4 x +2cot^2 x −1)))dx= [t=cot x → dx=−(dt/(csc^2 x))] =−(1/2)∫((t(t^4 −4t^2 +3))/(3t^4 +2t^2 −1))dt=−(1/6)∫tdt+(1/3)∫((t(7t^2 −5))/(3t^4 +2t^2 −1))dt −(1/6)∫tdt=−(t^2 /(12))=−((cot^2 x)/(12)) (1/3)∫((t(7t^2 −5))/(3t^4 +2t^2 −1))dt=∫(t/(t^2 +1))dt−(2/3)∫(t/(3t^2 −1))dt ∫(t/(t^2 +1))dt=(1/2)ln (t^2 +1) =(1/2)ln (1+cot^2 x) =ln ∣csc x∣ −(2/3)∫(t/(3t^2 −1))dt=−(1/9)ln (3t^2 −1)=−(1/9)ln ∣1−3cot^2 x∣ ∫cot x cot 2x cot 3x dx= =ln ∣csc x∣ −(1/9)ln ∣1−3cot^2 x∣ −((cot^2 x)/(12))+C

$$\int\mathrm{cot}\:{x}\:\mathrm{cot}\:\mathrm{2}{x}\:\mathrm{cot}\:\mathrm{3}{x}\:{dx}= \\ $$$$=\int\left(\mathrm{cot}\:{x}\:×\frac{\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{2cos}\:{x}\:\mathrm{sin}\:{x}}×\frac{\mathrm{cos}\:{x}\:\left(\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{3sin}^{\mathrm{2}} \:{x}\right)}{\mathrm{sin}\:{x}\:\left(\mathrm{3cos}^{\mathrm{2}} \:{x}\:−\mathrm{sin}^{\mathrm{2}} \:{x}\right.}\right){dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{cos}\:{x}\:=\frac{\mathrm{cot}\:{x}}{\mathrm{csc}\:{x}};\:\mathrm{sin}\:{x}\:=\frac{\mathrm{1}}{\mathrm{csc}\:{x}};\:\mathrm{csc}^{\mathrm{2}} \:{x}\:=\mathrm{1}+\mathrm{cot}^{\mathrm{2}} \:{x}\right] \\ $$$$=\int\frac{\mathrm{csc}^{\mathrm{2}} \:{x}\:\mathrm{cot}\:{x}\:\left(\mathrm{cot}^{\mathrm{4}} \:{x}\:−\mathrm{4cot}^{\mathrm{2}} \:{x}\:+\mathrm{3}\right)}{\mathrm{2}\left(\mathrm{3cot}^{\mathrm{4}} \:{x}\:+\mathrm{2cot}^{\mathrm{2}} \:{x}\:−\mathrm{1}\right)}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{cot}\:{x}\:\rightarrow\:{dx}=−\frac{{dt}}{\mathrm{csc}^{\mathrm{2}} \:{x}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{t}\left({t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} +\mathrm{3}\right)}{\mathrm{3}{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}}{dt}=−\frac{\mathrm{1}}{\mathrm{6}}\int{tdt}+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{t}\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{5}\right)}{\mathrm{3}{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$ \\ $$$$\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{6}}\int{tdt}=−\frac{{t}^{\mathrm{2}} }{\mathrm{12}}=−\frac{\mathrm{cot}^{\mathrm{2}} \:{x}}{\mathrm{12}} \\ $$$$ \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{t}\left(\mathrm{7}{t}^{\mathrm{2}} −\mathrm{5}\right)}{\mathrm{3}{t}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}}{dt}=\int\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{t}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}{dt} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{1}+\mathrm{cot}^{\mathrm{2}} \:{x}\right)\:=\mathrm{ln}\:\mid\mathrm{csc}\:{x}\mid \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{t}}{\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}}{dt}=−\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:\left(\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:\mid\mathrm{1}−\mathrm{3cot}^{\mathrm{2}} \:{x}\mid \\ $$$$ \\ $$$$\int\mathrm{cot}\:{x}\:\mathrm{cot}\:\mathrm{2}{x}\:\mathrm{cot}\:\mathrm{3}{x}\:{dx}= \\ $$$$=\mathrm{ln}\:\mid\mathrm{csc}\:{x}\mid\:−\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:\mid\mathrm{1}−\mathrm{3cot}^{\mathrm{2}} \:{x}\mid\:−\frac{\mathrm{cot}^{\mathrm{2}} \:{x}}{\mathrm{12}}+{C} \\ $$

Commented by Meritguide1234 last updated on 23/Oct/18

good approach

$${good}\:{approach} \\ $$

Commented by Meritguide1234 last updated on 23/Oct/18

Commented by Meritguide1234 last updated on 23/Oct/18

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