Question Number 49939 by turbo msup by abdo last updated on 12/Dec/18
$${find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sinx}\:{ln}\left(\mathrm{1}+{x}\right)\:{dx} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18
Commented by tanmay.chaudhury50@gmail.com last updated on 12/Dec/18
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sinxln}\left(\mathrm{1}+{x}\right){dx}\approx\left(\mathrm{0}.\mathrm{2}×\mathrm{0}.\mathrm{2}\right)×{nos}\:{smallest}\:{square} \\ $$
Commented by mathmax by abdo last updated on 03/Nov/19
![let determine a approximate value let A=∫_0 ^(π/2) sinx ln(1+x)dx by parts A =[−cosx ln(1+x)]_0 ^(π/2) −∫_0 ^(π/2) (−cosx)(dx/(1+x)) =∫_0 ^(π/2) ((cosx)/(1+x))dx we have cosx =Σ_(n=0) ^∞ (((−1)^n x^(2n) )/((2n)!)) =1−(x^2 /2) +(x^2 /(24))− ⇒1−(x^2 /2)≤cosx ≤1 ⇒((1−(x^2 /2))/(x+1))≤((cosx)/(x+1))≤(1/(x+1)) ⇒ ∫_0 ^(π/2) ((2−x^2 )/(2(x+1)))dx ≤∫_0 ^(π/2) ((cosx)/(x+1))dx ≤∫_0 ^1 (dx/(x+1)) ∫_0 ^1 (dx/(x+1)) =[ln(x+1)]_0 ^1 =ln(2) ∫_0 ^(π/2) ((2−x^2 )/(2(x+1)))dx =_(x+1=t) ∫_1 ^(1+(π/2)) ((2−(t−1)^2 )/(2t))dt =∫_1 ^(1+(π/2)) ((2−t^2 +2t−1)/t)dt =∫_1 ^(1+(π/2)) ((1−t^2 +2t)/t)dt =∫_1 ^(1+(π/2)) (dt/t) −∫_1 ^(1+(π/2)) tdt +π =ln(1+(π/2))−[(t^2 /2)]_1 ^(1+(π/2)) +π =ln(1+(π/2))−(1/2)((1+(π/2))^2 −1)+π ⇒ ln(1+(π/2))−(1/2)(π+(π^2 /4))+π≤A≤ln(2) ⇒ ln(1+(π/2))+(π/2)−(π^2 /8) ≤A ≤ln(2) and we can take v_0 =(1/2){ ln(1+(π/2))+(π/2)−(π^2 /8) +ln(2)} as aporoximate value for A](Q72809.png)
$${let}\:{determine}\:{a}\:{approximate}\:{value}\:{let}\:{A}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sinx}\:{ln}\left(\mathrm{1}+{x}\right){dx} \\ $$$${by}\:{parts}\:{A}\:=\left[−{cosx}\:{ln}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(−{cosx}\right)\frac{{dx}}{\mathrm{1}+{x}} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cosx}}{\mathrm{1}+{x}}{dx}\:\:\:{we}\:{have}\:\:{cosx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{24}}−\:\Rightarrow\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\leqslant{cosx}\:\leqslant\mathrm{1}\:\Rightarrow\frac{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}{{x}+\mathrm{1}}\leqslant\frac{{cosx}}{{x}+\mathrm{1}}\leqslant\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{2}−{x}^{\mathrm{2}} }{\mathrm{2}\left({x}+\mathrm{1}\right)}{dx}\:\leqslant\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cosx}}{{x}+\mathrm{1}}{dx}\:\leqslant\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}+\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}+\mathrm{1}}\:=\left[{ln}\left({x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} ={ln}\left(\mathrm{2}\right) \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{2}−{x}^{\mathrm{2}} }{\mathrm{2}\left({x}+\mathrm{1}\right)}{dx}\:=_{{x}+\mathrm{1}={t}} \:\:\:\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}{t}}{dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{2}−{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}}{{t}}{dt}\:=\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \:\:\frac{\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{2}{t}}{{t}}{dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{{t}}\:−\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} {tdt}\:+\pi\:={ln}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)−\left[\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \:+\pi \\ $$$$={ln}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}\right)+\pi\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\pi+\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\right)+\pi\leqslant{A}\leqslant{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\leqslant{A}\:\leqslant{ln}\left(\mathrm{2}\right)\:\:{and}\:{we}\:{can}\:{take} \\ $$$${v}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\:{ln}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)+\frac{\pi}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:+{ln}\left(\mathrm{2}\right)\right\}\:{as}\:{aporoximate}\:{value}\:{for}\:{A} \\ $$
Commented by mathmax by abdo last updated on 03/Nov/19
$${error}\:{of}\:{typo}\:{at}\:{line}\:\:\mathrm{8}\:\: \\ $$$$...=\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2}−{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}}{\mathrm{2}{t}}{dt}\:=\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−{t}^{\mathrm{2}} \:+\mathrm{2}{t}}{\mathrm{2}{t}}{dt} \\ $$$$=\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{2}{t}}−\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \:\frac{{t}}{\mathrm{2}}{dt}\:+\int_{\mathrm{1}} ^{\mathrm{1}+\frac{\pi}{\mathrm{2}}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left\{\:\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}\right\}+\frac{\pi}{\mathrm{2}}\:=..... \\ $$