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Question Number 156008 by zainaltanjung last updated on 07/Oct/21

Tekhnic Integration by part 1) Find ∫x.sec^2 x dx 2) Find ∫x.e^(2x) dx 3) Find ∫ln x dx 4) Find ∫x^2 .e^(2x) dx 5) Find ∫e^x cos x dx

$$\mathrm{Tekhnic}\:\:\mathrm{Integration}\:\mathrm{by}\:\mathrm{part} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Find}\:\int\mathrm{x}.\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Find}\:\int\mathrm{x}.\mathrm{e}^{\mathrm{2x}} \:\mathrm{dx} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{Find}\:\int\mathrm{ln}\:\:\mathrm{x}\:\mathrm{dx} \\ $$$$\left.\mathrm{4}\right)\:\:\mathrm{Find}\:\int\mathrm{x}^{\mathrm{2}} .\mathrm{e}^{\mathrm{2x}} \:\mathrm{dx} \\ $$$$\left.\mathrm{5}\right)\:\:\mathrm{Find}\:\int\mathrm{e}^{\mathrm{x}} \:\mathrm{cos}\:\mathrm{x}\:\mathrm{dx} \\ $$$$\: \\ $$$$ \\ $$

Commented by SANOGO last updated on 07/Oct/21

cool

$${cool} \\ $$

Commented by puissant last updated on 07/Oct/21

1) A=∫x sec^2 x dx = xtanx−∫tanx =xtanx−(−ln∣cosx∣)+C ⇒ A= xtanx+ln∣cosx∣+C..

$$\left.\mathrm{1}\right) \\ $$$${A}=\int{x}\:{sec}^{\mathrm{2}} {x}\:{dx}\:=\:{xtanx}−\int{tanx} \\ $$$$={xtanx}−\left(−{ln}\mid{cosx}\mid\right)+{C} \\ $$$$\Rightarrow\:{A}=\:{xtanx}+{ln}\mid{cosx}\mid+{C}.. \\ $$

Commented by puissant last updated on 07/Oct/21

2) S=∫x e^(2x) dx = (1/2)x e^(2x) −(1/2)∫e^(2x) dx =(1/2)x e^(2x) −(1/4)e^(2x) +C ⇒ S= (1/2){x e^(2x) −(1/2)e^(2x) }+C

$$\left.\mathrm{2}\right) \\ $$$${S}=\int{x}\:{e}^{\mathrm{2}{x}} {dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:{e}^{\mathrm{2}{x}} −\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{\mathrm{2}{x}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}\:{e}^{\mathrm{2}{x}} −\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{2}{x}} +{C} \\ $$$$\Rightarrow\:{S}=\:\frac{\mathrm{1}}{\mathrm{2}}\left\{{x}\:{e}^{\mathrm{2}{x}} −\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} \right\}+{C} \\ $$

Commented by SANOGO last updated on 07/Oct/21

3) ∫lnx=xlnx−x +c

$$\left.\mathrm{3}\right)\:\:\int{lnx}={xlnx}−{x}\:+{c} \\ $$

Commented by puissant last updated on 07/Oct/21

3) I=∫lnx dx { ((u=lnx)),((v′=1)) :} ⇒ { ((u′=(1/x))),((v=x)) :} ⇒ I = xlnx−∫1dx ⇒ I = xlnx − x + C

$$\left.\mathrm{3}\right) \\ $$$${I}=\int{lnx}\:{dx}\: \\ $$$$\begin{cases}{{u}={lnx}}\\{{v}'=\mathrm{1}}\end{cases}\:\:\Rightarrow\:\begin{cases}{{u}'=\frac{\mathrm{1}}{{x}}}\\{{v}={x}}\end{cases} \\ $$$$\Rightarrow\:{I}\:=\:{xlnx}−\int\mathrm{1}{dx} \\ $$$$\Rightarrow\:{I}\:=\:{xlnx}\:−\:{x}\:+\:{C} \\ $$

Commented by puissant last updated on 07/Oct/21

4) J=∫x^2 e^(2x) dx { ((u=x^2 )),((v′=e^(2x) )) :} ⇒ { ((u′=2x)),((v=(1/2)e^(2x) )) :} ⇒ J = (x^2 /2)e^(2x) −∫xe^(2x) dx ⇒ J=(x^2 /2)e^(2x) −(x/2)e^(2x) +(1/4)e^(2x) +C

$$\left.\mathrm{4}\right) \\ $$$${J}=\int{x}^{\mathrm{2}} {e}^{\mathrm{2}{x}} {dx}\: \\ $$$$\begin{cases}{{u}={x}^{\mathrm{2}} }\\{{v}'={e}^{\mathrm{2}{x}} }\end{cases}\:\:\:\Rightarrow\:\:\begin{cases}{{u}'=\mathrm{2}{x}}\\{{v}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} }\end{cases} \\ $$$$\Rightarrow\:{J}\:=\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{e}^{\mathrm{2}{x}} −\int{xe}^{\mathrm{2}{x}} {dx} \\ $$$$\Rightarrow\:{J}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{e}^{\mathrm{2}{x}} −\frac{{x}}{\mathrm{2}}{e}^{\mathrm{2}{x}} +\frac{\mathrm{1}}{\mathrm{4}}{e}^{\mathrm{2}{x}} +{C} \\ $$

Commented by puissant last updated on 07/Oct/21

5) D = ∫e^x cosx dx { ((u=cosx)),((v′=e^x )) :} ⇒ { ((u′=−sinx)),((v=e^x )) :} ⇒ D = e^x cosx+∫e^x sinx dx { ((u=sinx)),((v′=e^x )) :} ⇒ { ((u′=cosx)),((v=e^x )) :} ⇒ D = e^x cosx+e^x sinx−∫e^x cosxdx ⇒ D = e^x (cosx+sinx)−D ⇒ 2D=e^x (cosx+sinx) ⇒ D=(e^x /2)(cosx+sinx)+C..

$$\left.\mathrm{5}\right) \\ $$$${D}\:=\:\int{e}^{{x}} {cosx}\:{dx} \\ $$$$\begin{cases}{{u}={cosx}}\\{{v}'={e}^{{x}} }\end{cases}\:\:\Rightarrow\:\begin{cases}{{u}'=−{sinx}}\\{{v}={e}^{{x}} }\end{cases} \\ $$$$\Rightarrow\:{D}\:=\:{e}^{{x}} {cosx}+\int{e}^{{x}} {sinx}\:{dx} \\ $$$$\begin{cases}{{u}={sinx}}\\{{v}'={e}^{{x}} }\end{cases}\:\:\Rightarrow\:\:\begin{cases}{{u}'={cosx}}\\{{v}={e}^{{x}} }\end{cases} \\ $$$$\Rightarrow\:{D}\:=\:{e}^{{x}} {cosx}+{e}^{{x}} {sinx}−\int{e}^{{x}} {cosxdx} \\ $$$$\Rightarrow\:{D}\:=\:{e}^{{x}} \left({cosx}+{sinx}\right)−{D} \\ $$$$\Rightarrow\:\mathrm{2}{D}={e}^{{x}} \left({cosx}+{sinx}\right) \\ $$$$\Rightarrow\:{D}=\frac{{e}^{{x}} }{\mathrm{2}}\left({cosx}+{sinx}\right)+{C}.. \\ $$

Commented by tabata last updated on 08/Oct/21

3 ) with out by part ∫ lnx dx = ∫ ( lnx + 1 − 1 )dx = ∫ (( lnx + 1 ) − 1) dx = ∫ d ( x lnx ) − ∫ dx = x lnx − x + C □ M

$$\left.\mathrm{3}\:\right)\:\boldsymbol{{with}}\:\boldsymbol{{out}}\:\boldsymbol{{by}}\:\boldsymbol{{part}}\: \\ $$$$ \\ $$$$\int\:\boldsymbol{{lnx}}\:\boldsymbol{{dx}}\:=\:\int\:\left(\:\boldsymbol{{lnx}}\:+\:\mathrm{1}\:−\:\mathrm{1}\:\right)\boldsymbol{{dx}}\: \\ $$$$ \\ $$$$=\:\int\:\left(\left(\:\boldsymbol{{lnx}}\:+\:\mathrm{1}\:\right)\:−\:\mathrm{1}\right)\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\:=\:\int\:\boldsymbol{{d}}\:\left(\:\boldsymbol{{x}}\:\boldsymbol{{lnx}}\:\right)\:−\:\int\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$=\:\boldsymbol{{x}}\:\boldsymbol{{lnx}}\:−\:\boldsymbol{{x}}\:+\:\boldsymbol{{C}} \\ $$$$ \\ $$$$\square\:\boldsymbol{{M}} \\ $$

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