Question Number 152572 by rexford last updated on 29/Aug/21
$$\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \frac{\mathrm{1}+{cosx}}{\mathrm{3}+\mathrm{2}{sinx}}{dx} \\ $$$${please},{help}\:{me} \\ $$
Answered by Ar Brandon last updated on 29/Aug/21
![I=∫_(−(π/2)) ^(π/2) ((1+cosx)/(3+2sinx))dx, t=tan(x/2)⇒2dt=(1+t^2 )dx =∫_(−1) ^1 ((1+((1−t^2 )/(1+t^2 )))/(3+2((2t)/(1+t^2 ))))∙(2/(1+t^2 ))dt=4∫_(−1) ^1 (dt/((1+t^2 )(3t^2 +4t+3))) (1/((1+t^2 )(3t^2 +4t+3)))=((at+b)/(t^2 +1))+((ct+d)/(3t^2 +4t+3)) =(((at+b)(3t^2 +4t+3)+(ct+d)(t^2 +1))/) 3a+c=0, 3b+d=1, 4a+3b+d=0⇒a=−(1/4), c=(3/4), 3a+4b+c=0⇒b=0, d=1 I=4∫_(−1) ^1 (−(t/(4(t^2 +1)))+((3t+4)/(4(3t^2 +4t+3))))dt =4[−((ln(t^2 +1))/8)+((ln(3t^2 +4t+3))/8)]_(−1) ^1 +4∫_(−1) ^1 (dt/(3t^2 +4t+3)) =ln(√5)+(4/3)∫_(−1) ^1 (dt/(t^2 +(4/3)t+1))=ln(√5)+(4/3)∙(3/2)∙[arctan(((3t+4)/2))]_(−1) ^1 =ln(√5)+2(arctan(7)−arctan((1/2)))](Q152582.png)
$${I}=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\mathrm{cos}{x}}{\mathrm{3}+\mathrm{2sin}{x}}{dx},\:{t}=\mathrm{tan}\frac{{x}}{\mathrm{2}}\Rightarrow\mathrm{2}{dt}=\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dx} \\ $$$$\:\:=\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{3}+\mathrm{2}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }}\centerdot\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{4}\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{3}\right)} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{3}\right)}=\frac{{at}+{b}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{{ct}+{d}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{3}} \\ $$$$=\frac{\left({at}+{b}\right)\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{3}\right)+\left({ct}+{d}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{} \\ $$$$\mathrm{3}{a}+{c}=\mathrm{0},\:\mathrm{3}{b}+{d}=\mathrm{1},\:\mathrm{4}{a}+\mathrm{3}{b}+{d}=\mathrm{0}\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{4}},\:{c}=\frac{\mathrm{3}}{\mathrm{4}},\: \\ $$$$\mathrm{3}{a}+\mathrm{4}{b}+{c}=\mathrm{0}\Rightarrow{b}=\mathrm{0},\:{d}=\mathrm{1} \\ $$$${I}=\mathrm{4}\int_{−\mathrm{1}} ^{\mathrm{1}} \left(−\frac{{t}}{\mathrm{4}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{3}{t}+\mathrm{4}}{\mathrm{4}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{3}\right)}\right){dt} \\ $$$$\:\:=\mathrm{4}\left[−\frac{\mathrm{ln}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{8}}+\frac{\mathrm{ln}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{3}\right)}{\mathrm{8}}\right]_{−\mathrm{1}} ^{\mathrm{1}} +\mathrm{4}\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{3}} \\ $$$$\:\:=\mathrm{ln}\sqrt{\mathrm{5}}+\frac{\mathrm{4}}{\mathrm{3}}\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{3}}{t}+\mathrm{1}}=\mathrm{ln}\sqrt{\mathrm{5}}+\frac{\mathrm{4}}{\mathrm{3}}\centerdot\frac{\mathrm{3}}{\mathrm{2}}\centerdot\left[\mathrm{arctan}\left(\frac{\mathrm{3}{t}+\mathrm{4}}{\mathrm{2}}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} \\ $$$$\:\:=\mathrm{ln}\sqrt{\mathrm{5}}+\mathrm{2}\left(\mathrm{arctan}\left(\mathrm{7}\right)−\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$
Commented by SANOGO last updated on 30/Aug/21
Commented by puissant last updated on 30/Aug/21
![lim_(n→∞) (1/n)Σ_(k=1) ^n ln(1+(k^2 /n^2 )) =lim_(n→∞) ((b−a)/n)f(a+k((b−a)/n)) =∫_0 ^1 ln(1+x^2 )dx=K { ((u=ln(1+x^2 ))),((v′=1)) :} ⇒ { ((u′=((2x)/(1+x^2 )))),((v=x)) :} K=[xln(1+x^2 )]_0 ^1 −2∫_0 ^1 (x^2 /(1+x^2 ))dx =ln2−2∫_0 ^1 ((x^2 +1)/(1+x^2 ))dx+2∫_0 ^1 (1/(1+x^2 ))dx =ln2−2[x]_0 ^1 +2[arctanx]_0 ^1 =ln2−2+(π/2) K=lnQ ⇒ Q=e^K .. ⇒ Q=e^(ln2−2+(π/2)) = 2(e^(π/2) /e^2 )..](Q152586.png)
$${lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{1}+\frac{{k}^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right) \\ $$$$={lim}_{{n}\rightarrow\infty} \frac{{b}−{a}}{{n}}{f}\left({a}+{k}\frac{{b}−{a}}{{n}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx}={K} \\ $$$$\begin{cases}{{u}={ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\\{{v}'=\mathrm{1}}\end{cases}\:\Rightarrow\:\begin{cases}{{u}'=\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }}\\{{v}={x}}\end{cases} \\ $$$${K}=\left[{xln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={ln}\mathrm{2}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$={ln}\mathrm{2}−\mathrm{2}\left[{x}\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{2}\left[{arctanx}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$={ln}\mathrm{2}−\mathrm{2}+\frac{\pi}{\mathrm{2}} \\ $$$${K}={lnQ}\:\Rightarrow\:{Q}={e}^{{K}} .. \\ $$$$\Rightarrow\:{Q}={e}^{{ln}\mathrm{2}−\mathrm{2}+\frac{\pi}{\mathrm{2}}} =\:\mathrm{2}\frac{{e}^{\frac{\pi}{\mathrm{2}}} }{{e}^{\mathrm{2}} }.. \\ $$
Commented by SANOGO last updated on 30/Aug/21
$${merci}\:{bien}\:{mo}\:{prof} \\ $$