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tanx-tan2x-tan3x-dx-Any-way-to-solve-this-without-the-use-of-partial-fractions-

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Question Number 105440 by Ar Brandon last updated on 28/Jul/20
∫tanx∙tan2x∙tan3x∙dx Any way to solve this without the use of partial fractions?
∫tanx⋅tan2x⋅tan3x⋅dxAnywaytosolvethiswithouttheuseofpartialfractions?
Commented by som(math1967) last updated on 29/Jul/20
tan3x=((tan2x+tanx)/(1−tan2xtanx)) ∴tan3x−tan2x−tanx=tan3xtan2xtanx ∴∫tan3xtan2xtanxdx =∫tan3xdx−∫tan2xdx−∫tanxdx= =(1/3)lnsec3x+(1/2)lncos2x+lncosx+C
tan3x=tan2x+tanx1−tan2xtanx∴tan3x−tan2x−tanx=tan3xtan2xtanx∴∫tan3xtan2xtanxdx=∫tan3xdx−∫tan2xdx−∫tanxdx==13lnsec3x+12lncos2x+lncosx+C
Commented by Ar Brandon last updated on 29/Jul/20
Wow! Thank you Sir. You understood me perfectly.
Wow!ThankyouSir.Youunderstoodmeperfectly.
Commented by som(math1967) last updated on 29/Jul/20
welcome
welcome
Answered by Ar Brandon last updated on 28/Jul/20
I=∫tanx∙tan2x∙tan3x∙dx =∫tanx∙((2tanx)/(1−tan^2 x))∙tan3x∙dx =2∫((tan^2 x)/(1−tan^2 x))∙tan3x∙dx=−2∫(((1−tan^2 x)−1)/(1−tan^2 x))∙tan3x∙dx =−2∫{tan3x−((tan3x)/(1−tan^2 x))}dx=((2ln∣cos3x∣)/3)+2∫((tan3x)/(1−tan^2 x))dx tan3x=((tan2x+tanx)/(1−tan2x∙tanx))=((((2tanx)/(1−tan^2 x))+tanx)/(1−((2tanx)/(1−tan^2 x))tanx)) =((2tanx+tanx∙(1−tan^2 x))/(1−tan^2 x−2tan^2 x))=((3tanx−tan^3 x)/(1−3tan^2 x)) =((tanx(tan^2 x−3))/(3tan^2 x−1))=(1/3)∙((tanx∙((3tan^2 x−1)−8))/(3tan^2 x−1)) =(1/3)∙{((tanx(3tan^2 x−1))/(3tan^2 x−1))−((8tanx)/(3tan^2 x−1))} =(1/3)∙{tanx−((8tanx)/(3tan^2 x−1))} ⇒((tan3x)/(1−tan^2 x))=(1/3){((tanx)/(1−tan^2 x))+((8tanx)/((3tan^2 x−1)(tan^2 x−1)))} Here is my try. What can be done next?
I=∫tanx⋅tan2x⋅tan3x⋅dx=∫tanx⋅2tanx1−tan2x⋅tan3x⋅dx=2∫tan2x1−tan2x⋅tan3x⋅dx=−2∫(1−tan2x)−11−tan2x⋅tan3x⋅dx=−2∫{tan3x−tan3x1−tan2x}dx=2ln∣cos3x∣3+2∫tan3x1−tan2xdxtan3x=tan2x+tanx1−tan2x⋅tanx=2tanx1−tan2x+tanx1−2tanx1−tan2xtanx=2tanx+tanx⋅(1−tan2x)1−tan2x−2tan2x=3tanx−tan3x1−3tan2x=tanx(tan2x−3)3tan2x−1=13⋅tanx⋅((3tan2x−1)−8)3tan2x−1=13⋅{tanx(3tan2x−1)3tan2x−1−8tanx3tan2x−1}=13⋅{tanx−8tanx3tan2x−1}⇒tan3x1−tan2x=13{tanx1−tan2x+8tanx(3tan2x−1)(tan2x−1)}Hereismytry.Whatcanbedonenext?
Commented by malwaan last updated on 29/Jul/20
try to use the short way sir
trytousetheshortwaysir

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