Întrebare o integrala

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?

Da, este corect, cu completarea:

$\int_{0}^{\pi}ln\left ( sint \right )dt=\int_{0}^{\frac{\pi}{2}}ln\left ( sint \right )dt+\int_{\frac{\pi}{2}}^{\pi}ln\left ( sint \right )dt$ si

$\int_{0}^{\frac{\pi}{2}}ln\left ( sint \right )dt=\int_{\frac{\pi}{2}}^{\pi}ln\left ( sint \right )dt$.