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Question-83331




Question Number 83331 by oyemi kemewari last updated on 01/Mar/20
Commented by jagoll last updated on 01/Mar/20
(13) ∫ sec x dx = ∫ ((sec x (sec x+tan x) dx)/(sec x+tan x))  = ∫ ((sec^2 x+sec x tan x dx)/(sec x + tan x))  = ∫ ((d(sec x+tan x))/(sec x+tan x))  = ln ∣sec x+tan x∣ + c
$$\left(\mathrm{13}\right)\:\int\:\mathrm{sec}\:\mathrm{x}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{sec}\:\mathrm{x}\:\left(\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}\right)\:\mathrm{dx}}{\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}} \\ $$$$=\:\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sec}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{x}\:\mathrm{dx}}{\mathrm{sec}\:\mathrm{x}\:+\:\mathrm{tan}\:\mathrm{x}} \\ $$$$=\:\int\:\frac{\mathrm{d}\left(\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}\right)}{\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}} \\ $$$$=\:\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}\mid\:+\:\mathrm{c} \\ $$
Commented by mr W last updated on 01/Mar/20
(1):  maximum =(√(Σ_(i=1) ^n a_i ^2 ))  see Q83341
$$\left(\mathrm{1}\right): \\ $$$${maximum}\:=\sqrt{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} ^{\mathrm{2}} } \\ $$$${see}\:{Q}\mathrm{83341} \\ $$
Commented by abdomathmax last updated on 01/Mar/20
15)we use the diffeomorphism(r,θ)→(x,y) /  x=rcosθ and y =rsinθ  we have x,y≥0 ⇒0≤θ≤(π/2)  ∫_0 ^∞ ∫_0 ^∞  e^(−(x^2 +y^2 )) dxdy =∫_0 ^∞  ∫_0 ^(π/2)  e^(−r^2 ) rdrdθ  =(π/2) ∫_0 ^∞ r e^(−r^2 ) dr =(π/2)[−(1/2)e^(−r^2 ) ]_0 ^(+∞)  =(π/4)
$$\left.\mathrm{15}\right){we}\:{use}\:{the}\:{diffeomorphism}\left({r},\theta\right)\rightarrow\left({x},{y}\right)\:/ \\ $$$${x}={rcos}\theta\:{and}\:{y}\:={rsin}\theta\:\:{we}\:{have}\:{x},{y}\geqslant\mathrm{0}\:\Rightarrow\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} {dxdy}\:=\int_{\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{e}^{−{r}^{\mathrm{2}} } {rdrd}\theta \\ $$$$=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\infty} {r}\:{e}^{−{r}^{\mathrm{2}} } {dr}\:=\frac{\pi}{\mathrm{2}}\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{r}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{+\infty} \:=\frac{\pi}{\mathrm{4}} \\ $$
Commented by abdomathmax last updated on 01/Mar/20
18) A =∫_0 ^a ∫_0 ^a   ((xdxdy)/( (√(x^2 +y^2 ))))   (we suppose a>0) we use  the diffeomorphism(r,θ)→(x,y)=(rcosθ,rsinθ) ⇒  0≤x,y≤a ⇒ 0≤x^2 ,y^2 ≤a^2  ⇒0≤x^2  +y^2  ≤2a^2  ⇒  0≤r^2 ≤2a^2  ⇒0≤r≤a(√2) ⇒  A=∫_0 ^(a(√2))  ∫_0 ^(π/2)  ((rcosθ rdrdθ)/r)  =∫_0 ^(a(√2)) rdr ∫_0 ^(π/2)  dθ =(π/2)[(r^2 /2)]_0 ^(a(√2)) =(π/2)(((2a^2 )/2)) =((πa^2 )/2)
$$\left.\mathrm{18}\right)\:{A}\:=\int_{\mathrm{0}} ^{{a}} \int_{\mathrm{0}} ^{{a}} \:\:\frac{{xdxdy}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\:\:\:\left({we}\:{suppose}\:{a}>\mathrm{0}\right)\:{we}\:{use} \\ $$$${the}\:{diffeomorphism}\left({r},\theta\right)\rightarrow\left({x},{y}\right)=\left({rcos}\theta,{rsin}\theta\right)\:\Rightarrow \\ $$$$\mathrm{0}\leqslant{x},{y}\leqslant{a}\:\Rightarrow\:\mathrm{0}\leqslant{x}^{\mathrm{2}} ,{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} \:\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:\leqslant\mathrm{2}{a}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{0}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{2}{a}^{\mathrm{2}} \:\Rightarrow\mathrm{0}\leqslant{r}\leqslant{a}\sqrt{\mathrm{2}}\:\Rightarrow \\ $$$${A}=\int_{\mathrm{0}} ^{{a}\sqrt{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{rcos}\theta\:{rdrd}\theta}{{r}} \\ $$$$=\int_{\mathrm{0}} ^{{a}\sqrt{\mathrm{2}}} {rdr}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{d}\theta\:=\frac{\pi}{\mathrm{2}}\left[\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{a}\sqrt{\mathrm{2}}} =\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{2}}\right)\:=\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by mathmax by abdo last updated on 01/Mar/20
∫  5^(lnx)  dx =I    changement lnx=t give  I=∫  5^t  e^t  dt =∫  e^t  ×e^(tln(5))  dt =∫ e^((1+ln5)t)  dt  =(1/(1+ln5))e^((1+ln5)t)  +c =(1/(1+ln5)) e^((1+ln5)lnx)  =(1/(1+ln5)).x.5^x  +c
$$\int\:\:\mathrm{5}^{{lnx}} \:{dx}\:={I}\:\:\:\:{changement}\:{lnx}={t}\:{give} \\ $$$${I}=\int\:\:\mathrm{5}^{{t}} \:{e}^{{t}} \:{dt}\:=\int\:\:{e}^{{t}} \:×{e}^{{tln}\left(\mathrm{5}\right)} \:{dt}\:=\int\:{e}^{\left(\mathrm{1}+{ln}\mathrm{5}\right){t}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}+{ln}\mathrm{5}}{e}^{\left(\mathrm{1}+{ln}\mathrm{5}\right){t}} \:+{c}\:=\frac{\mathrm{1}}{\mathrm{1}+{ln}\mathrm{5}}\:{e}^{\left(\mathrm{1}+{ln}\mathrm{5}\right){lnx}} \:=\frac{\mathrm{1}}{\mathrm{1}+{ln}\mathrm{5}}.{x}.\mathrm{5}^{{x}} \:+{c} \\ $$
Commented by mathmax by abdo last updated on 02/Mar/20
23) A_n =∫_0 ^∞ ∫_0 ^∞ e^(−xy)  sin(nx)dxdy =∫_0 ^∞ (∫_0 ^∞  e^(−yx ) sin(nx)dx)dy  ∫_0 ^∞  e^(−yx)  sin(nx) =Im(∫_0 ^∞  e^(−yx+inx) dx)  ∫_0 ^∞  e^((−y+in)x) dx =[(1/(−y+in)) e^((−y+in)x) ]_0 ^(+∞) =−(1/(−y+in)) =(1/(y−in))  =((y+in)/(y^2  +n^2 )) ⇒∫_0 ^∞  e^(−yx) sin(nx)dx =(n/(y^2 +n^2 )) ⇒  A_n =n∫_0 ^∞  (dy/(y^2  +n^2 )) =_(y=nu)    n∫_0 ^∞   ((ndu)/(n^2 (1+u^2 ))) =(π/2)  by fubinni we have  A_n =∫_0 ^∞   (∫_0 ^∞ e^(−xy) dy)sin(nx)dx  =∫_0 ^∞ [−(1/x)e^(−xy) ]_(y=0) ^∞  sin(nx)dx =∫_0 ^∞  ((sin(nx))/x)dx =(π/2)
$$\left.\mathrm{23}\right)\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} {e}^{−{xy}} \:{sin}\left({nx}\right){dxdy}\:=\int_{\mathrm{0}} ^{\infty} \left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{yx}\:} {sin}\left({nx}\right){dx}\right){dy} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{yx}} \:{sin}\left({nx}\right)\:={Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{yx}+{inx}} {dx}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{y}+{in}\right){x}} {dx}\:=\left[\frac{\mathrm{1}}{−{y}+{in}}\:{e}^{\left(−{y}+{in}\right){x}} \right]_{\mathrm{0}} ^{+\infty} =−\frac{\mathrm{1}}{−{y}+{in}}\:=\frac{\mathrm{1}}{{y}−{in}} \\ $$$$=\frac{{y}+{in}}{{y}^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:{e}^{−{yx}} {sin}\left({nx}\right){dx}\:=\frac{{n}}{{y}^{\mathrm{2}} +{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${A}_{{n}} ={n}\int_{\mathrm{0}} ^{\infty} \:\frac{{dy}}{{y}^{\mathrm{2}} \:+{n}^{\mathrm{2}} }\:=_{{y}={nu}} \:\:\:{n}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ndu}}{{n}^{\mathrm{2}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\pi}{\mathrm{2}}\:\:{by}\:{fubinni}\:{we}\:{have} \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{xy}} {dy}\right){sin}\left({nx}\right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left[−\frac{\mathrm{1}}{{x}}{e}^{−{xy}} \right]_{{y}=\mathrm{0}} ^{\infty} \:{sin}\left({nx}\right){dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{{x}}{dx}\:=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 02/Mar/20
I =∫_0 ^∞ ((1−e^(−ax) )/x) e^(−x)  dx⇒I =∫_0 ^∞ ((e^(−x) −e^(−(a+1)x) )/x)dx =ϕ(a)  ϕ^′ (a) =∫_0 ^∞  e^(−(a+1)x)  dx =[−(1/(a+1))e^(−(a+1)x) ]_0 ^(+∞) =(1/(a+1)) ⇒  ϕ(a) =k+ln(a+1)  we have ϕ(0)=0=k ⇒  I=ln(a+1) ⇒∫_0 ^∞ ((1−e^(−x) )/x)e^(−x)  dx =∫_0 ^∞  ((e^(−x) −e^(−2x) )/x)dx =ln(2)
$${I}\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{e}^{−{ax}} }{{x}}\:{e}^{−{x}} \:{dx}\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{x}} −{e}^{−\left({a}+\mathrm{1}\right){x}} }{{x}}{dx}\:=\varphi\left({a}\right) \\ $$$$\varphi^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({a}+\mathrm{1}\right){x}} \:{dx}\:=\left[−\frac{\mathrm{1}}{{a}+\mathrm{1}}{e}^{−\left({a}+\mathrm{1}\right){x}} \right]_{\mathrm{0}} ^{+\infty} =\frac{\mathrm{1}}{{a}+\mathrm{1}}\:\Rightarrow \\ $$$$\varphi\left({a}\right)\:={k}+{ln}\left({a}+\mathrm{1}\right)\:\:{we}\:{have}\:\varphi\left(\mathrm{0}\right)=\mathrm{0}={k}\:\Rightarrow \\ $$$${I}={ln}\left({a}+\mathrm{1}\right)\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{e}^{−{x}} }{{x}}{e}^{−{x}} \:{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{x}} −{e}^{−\mathrm{2}{x}} }{{x}}{dx}\:={ln}\left(\mathrm{2}\right) \\ $$
Answered by mind is power last updated on 02/Mar/20
13 )  ∫_0 ^(+∞) ((1−e^(−ax) )/x)e^(−x) dx  =∫_0 ^(+∞) ((e^(−x) −e^(−(a+1)x) )/x)dx=ln(((a+1)/1))=ln(a+1)  ∫_0 ^(+∞) ((−Σ_(k=1) ^(+∞) (((−ax)^k e^(−x) )/(k!)))/x)  =Σ_(k≥1) (−1)^(k+1) .(a^k /(k!)).∫_0 ^(+∞) x^(k−1) .e^(−x) dx  =Σ_(k≥1) (((−1)^(k+1) )/(k!))a^k .Γ(k)=Σ_(k≥1) (((−1)^(k+1) )/k)a^k =ln(1+a),∀a∈]0,1[
$$\left.\mathrm{13}\:\right)\:\:\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}−{e}^{−{ax}} }{{x}}{e}^{−{x}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{e}^{−{x}} −{e}^{−\left({a}+\mathrm{1}\right){x}} }{{x}}{dx}={ln}\left(\frac{{a}+\mathrm{1}}{\mathrm{1}}\right)={ln}\left({a}+\mathrm{1}\right) \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{−\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\left(−{ax}\right)^{{k}} {e}^{−{x}} }{{k}!}}{{x}} \\ $$$$=\underset{{k}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} .\frac{{a}^{{k}} }{{k}!}.\int_{\mathrm{0}} ^{+\infty} {x}^{{k}−\mathrm{1}} .{e}^{−{x}} {dx} \\ $$$$\left.=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}!}{a}^{{k}} .\Gamma\left({k}\right)=\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{k}}{a}^{{k}} ={ln}\left(\mathrm{1}+{a}\right),\forall{a}\in\right]\mathrm{0},\mathrm{1}\left[\right. \\ $$

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