Întrebare o integrala

Buna ziua
Sa se calculeze o integrala notata cu I si anume:
\[\int_0^{\dfrac{\pi}{2}}ln(sinx)dx\\ Autorul\ considera\ o\ integrala\ J\ egala\ cu:\\ \int_0^{\dfrac{\pi}{2}}ln(cosx)dx\\ Face\ substitutia\ x=\dfrac{\pi}{2}-tsi\ aduna\ cele\ doua\ integrale.\\ Rezultat\ posibil:\\ a)I=-\dfrac{\pi}{2}ln2;b)I=\dfrac{1}{2};c)I=\dfrac{\pi}{2}ln2;\\ d)I=-\dfrac{3}{2};e)I=-2\\ multumesc\]

Buna seara,

Fie \(I=\int_{0}^{\frac{\pi }{2}}ln\left ( sinx \right )dx\).

Faceti schimbarea de variabila \(t=\frac{\pi }{2}-x\Rightarrow dt=-dx\). Capetele de integrare sunt: \(x=0\Rightarrow t=\frac{\pi}{2}\) si \(x=\frac{\pi}{2}\Rightarrow t=0\), deci

\(I=-\int_{\frac{\pi}{2}}^{0}ln\left ( sin\left ( \frac{\pi }{2}-t \right ) \right )dt=\int_{0}^{\frac{\pi}{2}}ln\left ( cost \right )dt\), adica

\(I=\int_{0}^{\frac{\pi}{2}}ln\left ( cosx \right )dx\).

Atunci \(2I=\int_{0}^{\frac{\pi}{2}}ln\left ( sinx \right )dx+\int_{0}^{\frac{\pi}{2}}ln\left ( cosx \right )dx=\int_{0}^{\frac{\pi}{2}}ln\left ( sinx\cdot cosx \right )dx\)

\(=\int_{0}^{\frac{\pi}{2}}ln\left ( \frac{sin2x}{2} \right )dx=\int_{0}^{\frac{\pi}{2}}\left [ln\left ( sin2x \right )-ln2 \right ]dx=\)

\(=\int_{0}^{\frac{\pi}{2}}ln\left ( sin2x \right )dx-x\cdot ln2|_{0}^{\frac{\pi}{2}}=\int_{0}^{\frac{\pi}{2}}ln\left ( sin2x \right )dx-\frac{\pi}{2}ln2\).

Aveti \(2I=\int_{0}^{\frac{\pi}{2}}ln\left (sin2x \right )dx-\frac{\pi}{2}ln2\), deci \(I=-\frac{\pi}{2}ln2\) .

(pentru ca \(\int_{0}^{\frac{\pi }{2}}ln\left ( sin2x \right )dx=\int_{0}^{\frac{\pi}{2}}ln\left ( sinx \right )dx\) - verificati ).

Raspuns corect: \(a)\).

Buna seara
Va multumesc pentru atentie.
Numai ca eu am constatat ca:
\[\int_0^{\dfrac{\pi}{2}}ln(sin2x)=\int_0^{\dfrac{\pi}{2}}ln(2sinxcosx)dx=\\ \int_0^{\dfrac{\pi}{2}}ln2dx+\int_0^{\dfrac{\pi}{2}}lnsinxdx+\int_0^{\dfrac{\pi}{2}}lncosxdx=\\ =\dfrac{\pi}{2}ln2+2I\]
Ma puteti ajuta sa gasesc eroarea pe care o fac?
multumesc

Nu este nicio eroare.

Daca va uitati la postarea mea, scrie \(2I=\int_{0}^{\frac{\pi}{2}}ln\left (sin2x \right )dx-\frac{\pi}{2}ln2\).

Dumneavoastra scrieti ca va da \(\int_{0}^{\frac{\pi}{2}}ln\left (sin2x \right )dx= \frac{\pi}{2}ln2 +2I\).

Vedeti ceva diferenta intre cele doua egalitati?

Buna seara
Am inteles rezolvarea-va multumesc.
Pentru verificarea solicitata eu scriu:

\[\int_0^{\dfrac{\pi}{2}}ln(sin2x)dx=\dfrac{1}{2}\cdot\int_0^{\dfrac{\pi}{2}}ln(sin2x)d(2x)=\\ =\dfrac{1}{2}\cdot\int_0^{\pi}ln(sint)dt=\dfrac{1}{2}\cdot[\int_0^{\dfrac{\pi}{2}}ln(sint)dt+\int_0^{\dfrac{\pi}{2}}ln(sint)dt]=\\ =\dfrac{1}{2}I+\dfrac{1}{2}I=I\]



Este corect?