Question Number 219853 by hardmath last updated on 02/May/25
$$\mathrm{If}\:\:\:\mathrm{0}<\mathrm{a}\leqslant\mathrm{b} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \left(\mathrm{sinx}\right)^{\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} \:\centerdot\:\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{1}−\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} \:\mathrm{dx}\:\geqslant\:\frac{\mathrm{b}−\mathrm{a}}{\mathrm{2}} \\ $$
Answered by MrGaster last updated on 03/May/25
![∫_a ^( b) (sinx)^(2sin^2 x) (cos^2 x)^(cos^2 x) dx ∀x∈R,sin^2 x=t⇒cos^2 x=1−t f(t)=t^t (1−t)^(1−t) ,t∈(0,1) ln f(t)=t ln t+(1−t)ln(1−t) (d/dt)ln f(t)=ln t−ln(1−t) ln(t/(1−t))=0⇒t=(1/2) (d^2 /dt^2 )ln f(t)=(1/t)+(1/(1−t))>0 f((1/2))=((1/2))^(1/2) ((1/2))^(1/2) =(1/2) ∀t∈(0,1),f(t)≥(1/2) ∫_a ^( b) (sinx)^(2sin^2 x) ∙ (1−sin^2 x)^(1−sin^2 x) dx ≥ ((b−a)/2) [Q.E.D]](https://www.tinkutara.com/question/Q219895.png)
$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \left(\mathrm{sinx}\right)^{\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} \:\left(\mathrm{cos}^{\mathrm{2}} {x}\right)^{\mathrm{cos}^{\mathrm{2}} {x}} {dx} \\ $$$$\forall{x}\in\mathbb{R},\mathrm{sin}^{\mathrm{2}} {x}={t}\Rightarrow\mathrm{cos}^{\mathrm{2}} {x}=\mathrm{1}−{t} \\ $$$${f}\left({t}\right)={t}^{{t}} \left(\mathrm{1}−{t}\right)^{\mathrm{1}−{t}} ,{t}\in\left(\mathrm{0},\mathrm{1}\right) \\ $$$$\mathrm{ln}\:{f}\left({t}\right)={t}\:\mathrm{ln}\:{t}+\left(\mathrm{1}−{t}\right)\mathrm{ln}\left(\mathrm{1}−{t}\right) \\ $$$$\frac{{d}}{{dt}}\mathrm{ln}\:{f}\left({t}\right)=\mathrm{ln}\:{t}−\mathrm{ln}\left(\mathrm{1}−{t}\right) \\ $$$$\mathrm{ln}\frac{{t}}{\mathrm{1}−{t}}=\mathrm{0}\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{2}} }{{dt}^{\mathrm{2}} }\mathrm{ln}\:{f}\left({t}\right)=\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{\mathrm{1}−{t}}>\mathrm{0} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\forall{t}\in\left(\mathrm{0},\mathrm{1}\right),{f}\left({t}\right)\geq\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int_{\boldsymbol{\mathrm{a}}} ^{\:\boldsymbol{\mathrm{b}}} \left(\mathrm{sinx}\right)^{\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} \:\centerdot\:\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{1}−\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}} \:\mathrm{dx}\:\geqslant\:\frac{\mathrm{b}−\mathrm{a}}{\mathrm{2}} \\ $$$$\left[\mathrm{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$
Commented by hardmath last updated on 03/May/25
$$ \\ $$Perfect solutions as always, thank you very much, my dear professor
Commented by Nicholas666 last updated on 03/May/25
$${thanks} \\ $$