Question

$\displaystyle \sf \red{\int_{0}^{ \frac{ \pi}{2} } \ln(2 \cos(x)) ) \: dx}$​

Answer 1

$\displaystyle{ \sf\int_{0}^{\frac{\pi}{2}}\left( \ln{\left(2 \cos{\left(x \right)} \right)} \right)dx}$

First, calculate the corresponding indefinite integral:

$\displaystyle{ \sf\int{\ln{\left(2 \cos{\left(x \right)} \right)} d x}=\frac{i x^{2}}{2} - x \ln{\left(\cos{\left(x \right)} \right)} + x \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \frac{i \operatorname{Li}_{2}\left(- e^{2 i x}\right)}{2}}$

The interval of integration contains the point π/2, which is not in the domain of the integrand, so this is an improper integral of type 2.

To evaluate an integral on an interval, we use the Fundamental Theorem of Calculus. However, we need to use the limit, if an endpoint of the interval is special (is not in the domain of the function).

$\displaystyle \sf{\int_{0}^{\frac{\pi}{2}}\left( \ln{\left(2 \cos{\left(x \right)} \right)} \right)dx=\lim_{x \to \frac{\pi}{2}}\left(\frac{i x^{2}}{2} - x \ln{\left(\cos{\left(x \right)} \right)} + x \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \frac{i \operatorname{Li}_{2}\left(- e^{2 i x}\right)}{2}\right)-\left(\frac{i x^{2}}{2} - x \ln{\left(\cos{\left(x \right)} \right)} + x \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + \frac{i \operatorname{Li}_{2}\left(- e^{2 i x}\right)}{2}\right)|_{\left(x=0\right)}=\frac{i \pi^{2}}{4}}$

Answer 2

Step-by-step explanation:

We have,

$\displaystyle \sf \int_{0}^{ \frac{ \pi}{2} } \ln(2 \cos(x)) ) \: dx$

$\displaystyle = \sf \int_{0}^{ \frac{ \pi}{2} }\ln(2) \: dx + \int_{0}^{ \frac{ \pi}{2} }\ln( \cos(x)) \: dx$

$\displaystyle = \sf { \ln(2)\int_{0}^{ \frac{ \pi}{2} } \: dx + \int_{0}^{ \frac{ \pi}{2} }\ln( \cos(x)) \: dx}$

$\displaystyle = \sf { \ln(2) \cdot [x]_{0}^{ \frac{ \pi}{2} } + \int_{0}^{ \frac{ \pi}{2} }\ln( \cos(x)) \: dx}$

$\displaystyle = \tt { \dfrac{\pi}{2} \ln(2)+ I}$

$\bf{ \mapsto \blue{Solving \: \: I}: }$

$\displaystyle I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln( \cos(x)) \: dx}$

$\displaystyle \implies I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln \left( \cos \left( \dfrac{ \pi}{2} - x \right) \right) \: dx}$

$\displaystyle \implies I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln \left( \sin \left( x \right) \right) \: dx}$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln ( \sin ( x )) \: dx + \int_{0}^{ \frac{ \pi}{2} }\ln ( \cos ( x )) \: dx }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln ( \sin ( x )) + \ln ( \cos ( x )) \: dx }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln ( \sin ( x ) \cos ( x )) \: dx }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln \left( \dfrac {2\sin ( x ) \cos ( x )}{2} \right) \: dx }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln \left( \dfrac {\sin ( 2x ) }{2} \right) \: dx }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} }\ln \left( \sin ( 2x ) \right) - \ln(2)\: dx }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} } \ln \left( \sin ( 2x ) \right)dx - \int_{0}^{ \frac{ \pi}{2} }\ln(2)\: dx }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} } \ln \left( \sin ( 2x ) \right)dx - \ln(2)\int_{0}^{ \frac{ \pi}{2} }\: dx }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} } \ln \left( \sin ( 2x ) \right)dx - \ln(2) \: [x]_{0}^{ \frac{ \pi}{2} }}$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{ \pi}{2} } \ln \left( \sin ( 2x ) \right)dx - \dfrac{\pi}{2} \ln(2) }$

$\bf{put \: \: 2x = t} \\ \bf{ \implies2dx = dt}$

$\bf{ When \: \: x = 0 , \: \: t = 0 } \\ \bf{ When \: \: x = \dfrac{\pi}{2} , \: \: t = \pi }$

So,

$\displaystyle \implies 2I= \dfrac{1}{2} \sf { \int_{0}^{ \pi } \ln \left( \sin ( t ) \right)dt - \dfrac{\pi}{2} \ln(2) }$

Since,

$\sin(\pi - x) = \sin(x)$

So,

$\displaystyle \implies 2I= \dfrac{1}{2} \cdot2 \sf { \int_{0}^{ \frac{\pi }{2}} \ln \left( \sin ( t ) \right)dt - \dfrac{\pi}{2} \ln(2) }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{\pi }{2}} \ln \left( \sin \left( \frac{\pi}{2} - t \right) \right)dt - \dfrac{\pi}{2} \ln(2) }$

$\displaystyle \implies 2I= \sf { \int_{0}^{ \frac{\pi }{2}} \ln \left( \cos \left( t \right) \right)dt - \dfrac{\pi}{2} \ln(2) }$

$\displaystyle \implies 2I= \sf { I- \dfrac{\pi}{2} \ln(2) }$

$\displaystyle \implies I= \tt{ - \dfrac{\pi}{2} \ln(2) }$

So,

Required integral,

$\displaystyle = \tt { \dfrac{\pi}{2} \ln(2) - \dfrac{\pi}{2} \ln(2)}$

$\displaystyle = \tt { 0}$