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Question Number 111818 by bemath last updated on 05/Sep/20
(√(bemath )) (1)∫ ((cos x)/(2−cos x)) dx (2) f(x) = ∣x^3 ∣ ⇒ f ′(x) ?
$$\:\:\:\sqrt{{bemath}\:} \\ $$$$\left(\mathrm{1}\right)\int\:\frac{\mathrm{cos}\:{x}}{\mathrm{2}−\mathrm{cos}\:{x}}\:{dx}\: \\ $$$$\left(\mathrm{2}\right)\:{f}\left({x}\right)\:=\:\mid{x}^{\mathrm{3}} \mid\:\Rightarrow\:{f}\:'\left({x}\right)\:? \\ $$
Answered by bobhans last updated on 05/Sep/20
f(x) = ∣x^3 ∣ → { ((x^3 ; x >0)),((0 ; x = 0)),((−x^3 ; x < 0)) :} f ′(x) = { ((3x^2 ; x>0)),((0 ; x = 0)),((−3x^2 ; x < 0)) :} f ′(x) = 3x∣x∣
$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mid\mathrm{x}^{\mathrm{3}} \mid\:\rightarrow\begin{cases}{\mathrm{x}^{\mathrm{3}} \:;\:\mathrm{x}\:>\mathrm{0}}\\{\mathrm{0}\:;\:\mathrm{x}\:=\:\mathrm{0}}\\{−\mathrm{x}^{\mathrm{3}} \:;\:\mathrm{x}\:<\:\mathrm{0}}\end{cases} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\begin{cases}{\mathrm{3x}^{\mathrm{2}} \:;\:\mathrm{x}>\mathrm{0}}\\{\mathrm{0}\:;\:\mathrm{x}\:=\:\mathrm{0}}\\{−\mathrm{3x}^{\mathrm{2}} ;\:\mathrm{x}\:<\:\mathrm{0}}\end{cases} \\ $$$$\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\mathrm{3x}\mid\mathrm{x}\mid\: \\ $$
Commented by bemath last updated on 05/Sep/20
jooss
$${jooss} \\ $$
Answered by 1549442205PVT last updated on 05/Sep/20
1)Put tan(x/2)=t⇒dt=(1/2)(1+t^2 )dx cosx=((1−t^2 )/(1+t^2 )),2−cosx=((3t^2 +1)/(1+t^2 )).Hence, F=∫ ((cos x)/(2−cos x)) dx =2∫((1−t^2 )/((3t^2 +1)(1+t^2 )))dt =2∫((−(t^2 +1)+2)/((3t^2 +1)(1+t^2 )))dt=−2∫(dt/(3t^2 +1))+4∫(dt/((3t^2 +1)(t^2 +1))) =((−2)/3)∫ (dt/(t^2 +((1/( (√3))))^2 ))+2∫ ((1/(t^2 +(1/3)))−(1/(t^2 +1)))dt =(4/3)∫ (dt/(t^2 +((1/( (√3))))^2 ))−2∫(( dt)/(1+t^2 )) =(4/3).((√3)/2)tan^(−1) ((√3)t)−2tan^(−1) (t) 2)f(x)=∣x^3 ∣⇒f^2 (x)=x^6 .Derivative two sides by x we get 2f(x).f ′(x)=6x^5 ⇒f ′(x)=((3x^5 )/(f(x)))=((3x^5 )/(∣x^3 ∣))
$$\left.\mathrm{1}\right)\mathrm{Put}\:\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}=\mathrm{t}\Rightarrow\mathrm{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{cosx}=\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\mathrm{1}+\mathrm{t}^{\mathrm{2}} },\mathrm{2}−\mathrm{cosx}=\frac{\mathrm{3t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }.\mathrm{Hence}, \\ $$$$\mathrm{F}=\int\:\frac{\mathrm{cos}\:{x}}{\mathrm{2}−\mathrm{cos}\:{x}}\:{dx}\:=\mathrm{2}\int\frac{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }{\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt} \\ $$$$=\mathrm{2}\int\frac{−\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{2}}{\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt}=−\mathrm{2}\int\frac{\mathrm{dt}}{\mathrm{3t}^{\mathrm{2}} +\mathrm{1}}+\mathrm{4}\int\frac{\mathrm{dt}}{\left(\mathrm{3t}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\frac{−\mathrm{2}}{\mathrm{3}}\int\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }+\mathrm{2}\int\:\left(\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}\right)\mathrm{dt} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\int\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }−\mathrm{2}\int\frac{\:\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}\mathrm{t}\right)−\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{t}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{f}\left(\mathrm{x}\right)=\mid\mathrm{x}^{\mathrm{3}} \mid\Rightarrow\mathrm{f}\:^{\mathrm{2}} \left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{6}} .\mathrm{Derivative}\:\mathrm{two} \\ $$$$\mathrm{sides}\:\mathrm{by}\:\mathrm{x}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{2f}\left(\mathrm{x}\right).\mathrm{f}\:'\left(\mathrm{x}\right)=\mathrm{6x}^{\mathrm{5}} \Rightarrow\mathrm{f}\:'\left(\mathrm{x}\right)=\frac{\mathrm{3x}^{\mathrm{5}} }{\mathrm{f}\left(\mathrm{x}\right)}=\frac{\mathrm{3x}^{\mathrm{5}} }{\mid\mathrm{x}^{\mathrm{3}} \mid} \\ $$
Answered by Dwaipayan Shikari last updated on 05/Sep/20
−∫((−cosx)/(2−cosx))=−∫((2−cosx)/(2−cosx))−(2/(2−cosx))=−x+2.2∫(1/(2−((1−t^2 )/(1+t^2 )))).(1/(1+t^2 )) =−x+4∫(1/(3t^2 +1))dt (t=tan(x/2)) =−x+(4/3)∫(1/((t)^2 +((1/( (√3))))^2 ))dt =−x+(4/( (√3)))tan^(−1) ((√3)tan(x/2))+C
$$−\int\frac{−{cosx}}{\mathrm{2}−{cosx}}=−\int\frac{\mathrm{2}−{cosx}}{\mathrm{2}−{cosx}}−\frac{\mathrm{2}}{\mathrm{2}−{cosx}}=−{x}+\mathrm{2}.\mathrm{2}\int\frac{\mathrm{1}}{\mathrm{2}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=−{x}+\mathrm{4}\int\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:\:\:\:\left({t}={tan}\frac{{x}}{\mathrm{2}}\right) \\ $$$$=−{x}+\frac{\mathrm{4}}{\mathrm{3}}\int\frac{\mathrm{1}}{\left({t}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }{dt} \\ $$$$=−{x}+\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{3}}{tan}\frac{{x}}{\mathrm{2}}\right)+{C} \\ $$

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