Question Number 119681 by TANMAY PANACEA last updated on 26/Oct/20
![∫_0 ^π (√((1+cos2x)/2)) dx ∫_0 ^∞ [ne^(−x) ]dx](https://www.tinkutara.com/question/Q119681.png)
$$\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}{x}}{\mathrm{2}}}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} \left[{ne}^{−{x}} \right]{dx} \\ $$
Answered by bemath last updated on 26/Oct/20
![(√(((1+cos 2x)/2) )) = (√((1+2cos^2 x−1)/2)) = ∣ cos x ∣ → { ((cos x for 0≤x≤(π/2))),((−cos x for (π/2)≤x≤π)) :} I = ∫_0 ^(π/2) ∣cos x∣ dx + ∫_(π/2) ^π ∣cos x∣ dx I = sin x ]_0 ^(π/2) −(sin x)]_(π/2) ^π I=1−(0−1)=2](https://www.tinkutara.com/question/Q119682.png)
$$\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}\:}\:=\:\sqrt{\frac{\mathrm{1}+\mathrm{2cos}\:^{\mathrm{2}} {x}−\mathrm{1}}{\mathrm{2}}} \\ $$$$\:=\:\mid\:\mathrm{cos}\:{x}\:\mid\:\rightarrow\begin{cases}{\mathrm{cos}\:{x}\:{for}\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}}\\{−\mathrm{cos}\:{x}\:{for}\:\frac{\pi}{\mathrm{2}}\leqslant{x}\leqslant\pi}\end{cases} \\ $$$${I}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mid\mathrm{cos}\:{x}\mid\:{dx}\:+\:\underset{\frac{\pi}{\mathrm{2}}} {\overset{\pi} {\int}}\:\mid\mathrm{cos}\:{x}\mid\:{dx} \\ $$$$\left.{I}\left.\:=\:\mathrm{sin}\:{x}\:\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\left(\mathrm{sin}\:{x}\right)\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} \\ $$$${I}=\mathrm{1}−\left(\mathrm{0}−\mathrm{1}\right)=\mathrm{2} \\ $$
Commented by TANMAY PANACEA last updated on 26/Oct/20
$${excellent}\:{sir} \\ $$
Answered by mindispower last updated on 26/Oct/20
![∫_0 ^∞ [ne^(−x) ]dx=u_n t=ne^(−x) ⇒x=−ln((t/n))⇒dx=−(1/t)dt =∫_0 ^n (([t])/t)dt =Σ_(k=0) ^(n−1) ∫_k ^(k+1) (([t]dt)/t)=Σ_(k≤n−1) ∫_k ^(k+1) (k/t)dt =Σ_(k≤n−1) kln(1+(1/k))=Σ_(k≤n−1) (kln(k+1)−kln(k)) =Σ_(k≤n−1) ((k+1)ln(k+1)−kln(k)−ln(k+1))=u_n n=0,u_0 =0 n≥1 u_n =nln(n)−ln[Π_(k≤n−1) (k+1)]=nln(n)−ln(n!) u_n = { ((0 ,n=0)),((nln(n)−ln(n!) ,n≥1)) :}](https://www.tinkutara.com/question/Q119690.png)
$$\int_{\mathrm{0}} ^{\infty} \left[{ne}^{−{x}} \right]{dx}={u}_{{n}} \\ $$$${t}={ne}^{−{x}} \Rightarrow{x}=−{ln}\left(\frac{{t}}{{n}}\right)\Rightarrow{dx}=−\frac{\mathrm{1}}{{t}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{{n}} \frac{\left[{t}\right]}{{t}}{dt} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{{k}} ^{{k}+\mathrm{1}} \frac{\left[{t}\right]{dt}}{{t}}=\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}\int_{{k}} ^{{k}+\mathrm{1}} \frac{{k}}{{t}}{dt} \\ $$$$=\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}{kln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right)=\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}\left({kln}\left({k}+\mathrm{1}\right)−{kln}\left({k}\right)\right) \\ $$$$=\underset{{k}\leqslant{n}−\mathrm{1}} {\sum}\left(\left({k}+\mathrm{1}\right){ln}\left({k}+\mathrm{1}\right)−{kln}\left({k}\right)−{ln}\left({k}+\mathrm{1}\right)\right)={u}_{{n}} \\ $$$${n}=\mathrm{0},{u}_{\mathrm{0}} =\mathrm{0} \\ $$$${n}\geqslant\mathrm{1} \\ $$$${u}_{{n}} ={nln}\left({n}\right)−{ln}\left[\underset{{k}\leqslant{n}−\mathrm{1}} {\prod}\left({k}+\mathrm{1}\right)\right]={nln}\left({n}\right)−{ln}\left({n}!\right) \\ $$$${u}_{{n}} =\begin{cases}{\mathrm{0}\:\:,{n}=\mathrm{0}}\\{{nln}\left({n}\right)−{ln}\left({n}!\right)\:,{n}\geqslant\mathrm{1}}\end{cases} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by TANMAY PANACEA last updated on 26/Oct/20
$${thank}\:{you}\:{sir} \\ $$
Commented by mindispower last updated on 26/Oct/20
$${withe}\:{pleasur}\:{have}\:{a}\:{nice}\:{day} \\ $$
Answered by mathmax by abdo last updated on 26/Oct/20
![A =∫_0 ^π (√((1+cos(2x))/2))dx ⇒A =∫_0 ^π ∣cosx∣dx =∫_0 ^(π/2) cosx dx −∫_(π/2) ^π cosx dx =[sinx]_0 ^(π/2) −[sinx]_(π/2) ^π =1−(−1) =2 ⇒ A =2](https://www.tinkutara.com/question/Q119714.png)
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\pi} \sqrt{\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}}\mathrm{dx}\:\Rightarrow\mathrm{A}\:=\int_{\mathrm{0}} ^{\pi} \mid\mathrm{cosx}\mid\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cosx}\:\mathrm{dx}\:−\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \mathrm{cosx}\:\mathrm{dx}\:=\left[\mathrm{sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\left[\mathrm{sinx}\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} \\ $$$$=\mathrm{1}−\left(−\mathrm{1}\right)\:=\mathrm{2}\:\Rightarrow\:\mathrm{A}\:=\mathrm{2} \\ $$
Answered by mathmax by abdo last updated on 26/Oct/20
![I_n =∫_0 ^∞ [ne^(−x) ]dx we do the changement ne^(−x) =t ⇒e^(−x) =(t/n) ⇒−x=ln((t/n)) ⇒x =−ln((t/n))=ln(n)−ln(t) (we suppose n≠0) ⇒I_n =∫_n ^0 [t](−(dt/t)) =∫_0 ^n (([t])/t)dt =Σ_(k=0) ^(n−1) ∫_k ^(k+1) (k/t)dt =Σ_(k=0) ^(n−1) k{ln(k+1)−ln(k)} =Σ_(k=1) ^(n−1) kln(((k+1)/k)) =Σ_(k=1) ^(n−1) k ln(1+(1/k)) =ln(2)+2ln(1+(1/2))+3ln(1+(1/3))+...+(n−1)ln(1+(1/(n−1))) ifn=0 we get I_n =0](https://www.tinkutara.com/question/Q119719.png)
$$\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \left[\mathrm{ne}^{−\mathrm{x}} \right]\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{ne}^{−\mathrm{x}} \:=\mathrm{t}\:\Rightarrow\mathrm{e}^{−\mathrm{x}} \:=\frac{\mathrm{t}}{\mathrm{n}} \\ $$$$\Rightarrow−\mathrm{x}=\mathrm{ln}\left(\frac{\mathrm{t}}{\mathrm{n}}\right)\:\Rightarrow\mathrm{x}\:=−\mathrm{ln}\left(\frac{\mathrm{t}}{\mathrm{n}}\right)=\mathrm{ln}\left(\mathrm{n}\right)−\mathrm{ln}\left(\mathrm{t}\right)\:\left(\mathrm{we}\:\mathrm{suppose}\:\mathrm{n}\neq\mathrm{0}\right) \\ $$$$\Rightarrow\mathrm{I}_{\mathrm{n}} =\int_{\mathrm{n}} ^{\mathrm{0}} \left[\mathrm{t}\right]\left(−\frac{\mathrm{dt}}{\mathrm{t}}\right)\:=\int_{\mathrm{0}} ^{\mathrm{n}} \:\frac{\left[\mathrm{t}\right]}{\mathrm{t}}\mathrm{dt}\:\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \int_{\mathrm{k}} ^{\mathrm{k}+\mathrm{1}} \frac{\mathrm{k}}{\mathrm{t}}\mathrm{dt} \\ $$$$=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{n}−\mathrm{1}} \mathrm{k}\left\{\mathrm{ln}\left(\mathrm{k}+\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{k}\right)\right\} \\ $$$$=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \mathrm{kln}\left(\frac{\mathrm{k}+\mathrm{1}}{\mathrm{k}}\right)\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}−\mathrm{1}} \mathrm{k}\:\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{k}}\right) \\ $$$$=\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{3ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)+…+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}−\mathrm{1}}\right) \\ $$$$\mathrm{ifn}=\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{I}_{\mathrm{n}} =\mathrm{0} \\ $$$$ \\ $$