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Find-1-cos3x-cosx-dx-2-3-x-sinx-dx-3-0-1-x-e-x-dx-4-1-e-ln-2-x-dx-

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Question Number 201763 by hardmath last updated on 11/Dec/23
Find: 1. ∫ cos3x cosx dx = ? 2. ∫ 3^x sinx dx = ? 3. ∫_(0 ) ^( 1) x e^(−x) dx = ? 4. ∫_1 ^( e) ln^2 x dx = ?
$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\int\:\mathrm{cos3x}\:\mathrm{cosx}\:\mathrm{dx}\:=\:? \\ $$$$\mathrm{2}.\:\int\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:\mathrm{sinx}\:\mathrm{dx}\:=\:? \\ $$$$\mathrm{3}.\:\int_{\mathrm{0}\:} ^{\:\mathrm{1}} \:\mathrm{x}\:\mathrm{e}^{−\boldsymbol{\mathrm{x}}} \:\mathrm{dx}\:=\:? \\ $$$$\mathrm{4}.\:\int_{\mathrm{1}} ^{\:\boldsymbol{\mathrm{e}}} \:\mathrm{ln}^{\mathrm{2}} \:\mathrm{x}\:\mathrm{dx}\:=\:? \\ $$
Answered by Calculusboy last updated on 12/Dec/23
Solution: (3)∫_0 ^1 xe^(−x) dx (by using Bernoulli′s formula) u=x u′=1 u′′=0 and v=e^(−x) v_1 =−e^(−x) v_2 =e^(−x) v_3 =−e^(−x) I=uv_1 −u′v_2 +u′′v_3 −u′′′v_4 +∙∙∙ I=−xe^(−x) −e^(−x) I=−[xe^(−x) ]_0 ^1 −[e^(−x) ]_0 ^1 I=−[e^(−1) −0]−[e^(−1) −1] I=−e^(−1) −e^(−1) +1 I=1−(2/e)=((e−2)/e)✓
$$\boldsymbol{{Solution}}:\: \\ $$$$\left(\mathrm{3}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{xe}}^{−\boldsymbol{{x}}} \boldsymbol{{dx}}\:\:\:\left(\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{Bernoulli}}'\boldsymbol{{s}}\:\boldsymbol{{formula}}\right) \\ $$$$\boldsymbol{{u}}=\boldsymbol{{x}}\:\:\boldsymbol{{u}}'=\mathrm{1}\:\:\boldsymbol{{u}}''=\mathrm{0}\:\boldsymbol{{and}}\:\boldsymbol{{v}}=\boldsymbol{{e}}^{−\boldsymbol{{x}}} \:\:\boldsymbol{{v}}_{\mathrm{1}} =−\boldsymbol{{e}}^{−\boldsymbol{{x}}} \:\:\:\boldsymbol{{v}}_{\mathrm{2}} =\boldsymbol{{e}}^{−\boldsymbol{{x}}} \\ $$$$\boldsymbol{{v}}_{\mathrm{3}} =−\boldsymbol{{e}}^{−\boldsymbol{{x}}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{uv}}_{\mathrm{1}} −\boldsymbol{{u}}'\boldsymbol{{v}}_{\mathrm{2}} +\boldsymbol{{u}}''\boldsymbol{{v}}_{\mathrm{3}} −\boldsymbol{{u}}'''\boldsymbol{{v}}_{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$$\boldsymbol{{I}}=−\boldsymbol{{xe}}^{−\boldsymbol{{x}}} −\boldsymbol{{e}}^{−\boldsymbol{{x}}} \\ $$$$\boldsymbol{{I}}=−\left[\boldsymbol{{xe}}^{−\boldsymbol{{x}}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\left[\boldsymbol{{e}}^{−\boldsymbol{{x}}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\boldsymbol{{I}}=−\left[\boldsymbol{{e}}^{−\mathrm{1}} −\mathrm{0}\right]−\left[\boldsymbol{{e}}^{−\mathrm{1}} −\mathrm{1}\right] \\ $$$$\boldsymbol{{I}}=−\boldsymbol{{e}}^{−\mathrm{1}} −\boldsymbol{{e}}^{−\mathrm{1}} +\mathrm{1} \\ $$$$\boldsymbol{{I}}=\mathrm{1}−\frac{\mathrm{2}}{\boldsymbol{{e}}}=\frac{\boldsymbol{{e}}−\mathrm{2}}{\boldsymbol{{e}}}\checkmark \\ $$
Commented by hardmath last updated on 12/Dec/23
thank you ser
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{ser} \\ $$
Answered by Calculusboy last updated on 12/Dec/23
Solution: (2) ∫ 3^x sinx dx (by using IBP) let u=3^x du=3^x In(3)dx dv=sinx v=−cosx ∫udvdx=uv−∫vdudx let I=∫udvdx I=−3^x cosx+In(3)∫3^x cosx dx let I_1 =∫3^x cosxdx let u=3^x du=3^x In(3)dx dv=cosx v=sinx I_1 =uv−∫vdudx I_1 =3^x sinx−In(3)∫3^x sinxdx but I=∫3^x sinxdx I_1 =3^x sinx−In(3)I and I=−3^x cosx+In(3)I_1 I=−3^x cosx+In(3)[3^x sinx−In(3)I] I=−3^x cosx+3^x In(3)sinx−In^2 (3)I I+In^2 (3)I=3^x In(3)sinx−3^x cosx I=(3^x /( [1+In^2 (3)])){In(3)sinx−cosx}+C
$$\boldsymbol{{Solution}}: \\ $$$$\left(\mathrm{2}\right)\:\int\:\mathrm{3}^{\boldsymbol{{x}}} \:\boldsymbol{{sinx}}\:\boldsymbol{{dx}}\:\:\:\left(\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{IBP}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{u}}=\mathrm{3}^{\boldsymbol{{x}}} \:\:\boldsymbol{{du}}=\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{3}\right)\boldsymbol{{dx}}\:\:\:\boldsymbol{{dv}}=\boldsymbol{{sinx}}\:\:\boldsymbol{{v}}=−\boldsymbol{{cosx}} \\ $$$$\int\boldsymbol{{udvdx}}=\boldsymbol{{uv}}−\int\boldsymbol{{vdudx}}\:\:\:\boldsymbol{{let}}\:\boldsymbol{{I}}=\int\boldsymbol{{udvdx}} \\ $$$$\boldsymbol{{I}}=−\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{cosx}}+\boldsymbol{{In}}\left(\mathrm{3}\right)\int\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{cosx}}\:\boldsymbol{{dx}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{I}}_{\mathrm{1}} =\int\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{cosxdx}}\:\boldsymbol{{let}}\:\boldsymbol{{u}}=\mathrm{3}^{\boldsymbol{{x}}} \:\:\boldsymbol{{du}}=\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{3}\right)\boldsymbol{{dx}}\:\:\boldsymbol{{dv}}=\boldsymbol{{cosx}}\:\:\boldsymbol{{v}}=\boldsymbol{{sinx}} \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =\boldsymbol{{uv}}−\int\boldsymbol{{vdudx}} \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{sinx}}−\boldsymbol{{In}}\left(\mathrm{3}\right)\int\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{sinxdx}}\:\:\boldsymbol{{but}}\:\boldsymbol{{I}}=\int\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{sinxdx}} \\ $$$$\boldsymbol{{I}}_{\mathrm{1}} =\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{sinx}}−\boldsymbol{{In}}\left(\mathrm{3}\right)\boldsymbol{{I}}\:\:\boldsymbol{{and}}\:\boldsymbol{{I}}=−\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{cosx}}+\boldsymbol{{In}}\left(\mathrm{3}\right)\boldsymbol{{I}}_{\mathrm{1}} \\ $$$$\boldsymbol{{I}}=−\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{cosx}}+\boldsymbol{{In}}\left(\mathrm{3}\right)\left[\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{sinx}}−\boldsymbol{{In}}\left(\mathrm{3}\right)\boldsymbol{{I}}\right] \\ $$$$\boldsymbol{{I}}=−\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{cosx}}+\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{3}\right)\boldsymbol{{sinx}}−\boldsymbol{{In}}^{\mathrm{2}} \left(\mathrm{3}\right)\boldsymbol{{I}} \\ $$$$\boldsymbol{{I}}+\boldsymbol{{In}}^{\mathrm{2}} \left(\mathrm{3}\right)\boldsymbol{{I}}=\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{In}}\left(\mathrm{3}\right)\boldsymbol{{sinx}}−\mathrm{3}^{\boldsymbol{{x}}} \boldsymbol{{cosx}} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{3}^{\boldsymbol{{x}}} }{\:\left[\mathrm{1}+\boldsymbol{{In}}^{\mathrm{2}} \left(\mathrm{3}\right)\right]}\left\{\boldsymbol{{In}}\left(\mathrm{3}\right)\boldsymbol{{sinx}}−\boldsymbol{{cosx}}\right\}+\boldsymbol{{C}} \\ $$$$ \\ $$$$ \\ $$
Commented by hardmath last updated on 12/Dec/23
thank you ser, In = ln.?
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{ser}, \\ $$$$\boldsymbol{\mathrm{I}}\mathrm{n}\:=\:\mathrm{ln}.? \\ $$
Commented by Calculusboy last updated on 12/Dec/23
i mean it like this Ln
$$\boldsymbol{{i}}\:\boldsymbol{{mean}}\:\boldsymbol{{it}}\:\boldsymbol{{like}}\:\boldsymbol{{this}}\:\boldsymbol{{Ln}} \\ $$
Commented by mr W last updated on 12/Dec/23
you know the differen between I and l, but you wont make the difference and therefore cause confusions again and again. see Q201546
$${you}\:{know}\:{the}\:{differen}\:{between}\:{I}\:{and} \\ $$$${l},\:{but}\:{you}\:{wont}\:{make}\:{the}\:{difference} \\ $$$${and}\:{therefore}\:{cause}\:{confusions} \\ $$$${again}\:{and}\:{again}. \\ $$$${see}\:{Q}\mathrm{201546} \\ $$
Commented by hardmath last updated on 12/Dec/23
thank you dear professors
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professors} \\ $$
Commented by Calculusboy last updated on 12/Dec/23
am sorry for the confusing sir
$$\boldsymbol{{am}}\:\boldsymbol{{sorry}}\:\boldsymbol{{for}}\:\boldsymbol{{the}}\:\boldsymbol{{confusing}}\:\boldsymbol{{sir}} \\ $$
Answered by Calculusboy last updated on 12/Dec/23
Solution: (1) ∫cos3xcosxdx Recall that ∫cosaxcosbxdx=(1/2)∫[cos(a−b)x+cos(a+b)x]dx I=(1/2)∫[cos(3−1)x+cos(3+1)x]dx I=(1/2)∫[cos2x+cos4x]dx I=(1/2)[∫cos2xdx+∫cos4xdx] I=(1/2)[((sin2x)/2)+((sin4x)/4)]+C
$$\boldsymbol{{Solution}}: \\ $$$$\left(\mathrm{1}\right)\:\int\boldsymbol{{cos}}\mathrm{3}\boldsymbol{{xcosxdx}} \\ $$$$\boldsymbol{{Recall}}\:\boldsymbol{{that}}\:\int\boldsymbol{{cosaxcosbxdx}}=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\boldsymbol{{cos}}\left(\boldsymbol{{a}}−\boldsymbol{{b}}\right)\boldsymbol{{x}}+\boldsymbol{{cos}}\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)\boldsymbol{{x}}\right]\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\boldsymbol{{cos}}\left(\mathrm{3}−\mathrm{1}\right)\boldsymbol{{x}}+\boldsymbol{{cos}}\left(\mathrm{3}+\mathrm{1}\right)\boldsymbol{{x}}\right]\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}+\boldsymbol{{cos}}\mathrm{4}\boldsymbol{{x}}\right]\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{xdx}}+\int\boldsymbol{{cos}}\mathrm{4}\boldsymbol{{xdx}}\right] \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\mathrm{2}}+\frac{\boldsymbol{{sin}}\mathrm{4}\boldsymbol{{x}}}{\mathrm{4}}\right]+\boldsymbol{{C}} \\ $$$$ \\ $$
Answered by esmaeil last updated on 12/Dec/23
4: =Ω lnxdx=dv→v=xlnx−x lnx=u→(dx/x)=du →Ω=xln^2 x−xlnx−(∫^e _1 (lnx−1)dx)= xln^2 x−2xlnx+2x]^e _1 = e−2
$$\mathrm{4}:\:\:=\Omega\:\:{lnxdx}={dv}\rightarrow{v}={xlnx}−{x} \\ $$$$\:\:\:{lnx}={u}\rightarrow\frac{{dx}}{{x}}={du} \\ $$$$\rightarrow\Omega={xln}^{\mathrm{2}} {x}−{xlnx}−\left(\underset{\mathrm{1}} {\int}^{{e}} \left({lnx}−\mathrm{1}\right){dx}\right)= \\ $$$$\left.{xln}^{\mathrm{2}} {x}−\mathrm{2}{xlnx}+\mathrm{2}{x}\underset{\mathrm{1}} {\right]}^{{e}} = \\ $$$${e}−\mathrm{2} \\ $$
Answered by Calculusboy last updated on 12/Dec/23
Solution: (4) ∫_1 ^e Ln^2 x dx=∫_1 ^e (Lnx)^2 dx (by using u−sub) let p=Lnx xdp=dx x=e^p when x=e p=1 and when x=1 p=0 I=∫_0 ^1 p^2 ∙e^p dp (by using BF method) u=p^2 u′=2p u′′=2 u′′′=0 v=e^p v_1 =e^p v_2 =e^p v_3 =e^p v_4 =e^p I=uv_1 −u′v_2 +u′′v_3 −u′′′v_4 +∙∙∙ I=∣p^2 e^p ∣_0 ^1 −2∣pe^p ∣_0 ^1 +2∣e^p ∣_0 ^1 +C I=(e−0)−2(e−0)+2(e−1) I=e−2e+2e−2 I=e−2
$$\boldsymbol{{Solution}}: \\ $$$$\left(\mathrm{4}\right)\:\int_{\mathrm{1}} ^{\boldsymbol{{e}}} \boldsymbol{{Ln}}^{\mathrm{2}} \boldsymbol{{x}}\:\boldsymbol{{dx}}=\int_{\mathrm{1}} ^{\boldsymbol{{e}}} \left(\boldsymbol{{Lnx}}\right)^{\mathrm{2}} \boldsymbol{{dx}}\:\:\:\:\left(\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{u}}−\boldsymbol{{sub}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{p}}=\boldsymbol{{Lnx}}\:\:\:\boldsymbol{{xdp}}=\boldsymbol{{dx}}\:\:\boldsymbol{{x}}=\boldsymbol{{e}}^{\boldsymbol{{p}}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\boldsymbol{{e}}\:\boldsymbol{{p}}=\mathrm{1}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{1}\:\:\boldsymbol{{p}}=\mathrm{0} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\mathrm{1}} \boldsymbol{{p}}^{\mathrm{2}} \centerdot\boldsymbol{{e}}^{\boldsymbol{{p}}} \boldsymbol{{dp}}\:\:\left(\boldsymbol{{by}}\:\boldsymbol{{using}}\:\boldsymbol{{BF}}\:\boldsymbol{{method}}\right) \\ $$$$\boldsymbol{{u}}=\boldsymbol{{p}}^{\mathrm{2}} \:\:\boldsymbol{{u}}'=\mathrm{2}\boldsymbol{{p}}\:\:\boldsymbol{{u}}''=\mathrm{2}\:\:\:\boldsymbol{{u}}'''=\mathrm{0} \\ $$$$\boldsymbol{{v}}=\boldsymbol{{e}}^{\boldsymbol{{p}}} \:\:\boldsymbol{{v}}_{\mathrm{1}} =\boldsymbol{{e}}^{\boldsymbol{{p}}} \:\:\boldsymbol{{v}}_{\mathrm{2}} =\boldsymbol{{e}}^{\boldsymbol{{p}}} \:\:\boldsymbol{{v}}_{\mathrm{3}} =\boldsymbol{{e}}^{\boldsymbol{{p}}} \:\:\boldsymbol{{v}}_{\mathrm{4}} =\boldsymbol{{e}}^{\boldsymbol{{p}}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{uv}}_{\mathrm{1}} −\boldsymbol{{u}}'\boldsymbol{{v}}_{\mathrm{2}} +\boldsymbol{{u}}''\boldsymbol{{v}}_{\mathrm{3}} −\boldsymbol{{u}}'''\boldsymbol{{v}}_{\mathrm{4}} +\centerdot\centerdot\centerdot \\ $$$$\boldsymbol{{I}}=\mid\boldsymbol{{p}}^{\mathrm{2}} \boldsymbol{{e}}^{\boldsymbol{{p}}} \mid_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\mid\boldsymbol{{pe}}^{\boldsymbol{{p}}} \mid_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\mid\boldsymbol{{e}}^{\boldsymbol{{p}}} \mid_{\mathrm{0}} ^{\mathrm{1}} \:+\boldsymbol{{C}} \\ $$$$\boldsymbol{{I}}=\left(\boldsymbol{{e}}−\mathrm{0}\right)−\mathrm{2}\left(\boldsymbol{{e}}−\mathrm{0}\right)+\mathrm{2}\left(\boldsymbol{{e}}−\mathrm{1}\right) \\ $$$$\boldsymbol{{I}}=\boldsymbol{{e}}−\mathrm{2}\boldsymbol{{e}}+\mathrm{2}\boldsymbol{{e}}−\mathrm{2} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{e}}−\mathrm{2} \\ $$$$ \\ $$

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