For integration of $\log(\cos x)$, we generally consider the definite integral over the interval from 0 to $\pi/2$

To calculate the definite integral of $\log(\cos x)$ from (0) to $\pi/2$, we use a technique involving symmetry and the properties of logarithms. The integral is:

\[

\int_{0}^{\pi/2} \log(\cos x) \, dx

\]

Let

$I=\int_{0}^{\pi/2} \log(\cos x) \, dx$ –(1)

Now we know

$\int_{0}^{a} f(x) dx=\int_{0}^{a} f((a-x)) dx$

Therefore

$I=\int_{0}^{\pi/2} \log(\cos (\pi/2 – x) \, dx=\int_{0}^{\pi/2} \log(\sin x) \, dx $ –(2)

Adding (1) and (2)

$2I = \int_{0}^{\pi/2} \log(\sin x) + \log(\cos x) \, dx $

By properties of logarithms

$\log a + \log b = \log ab$

we have

$2I = \int_{0}^{\pi/2} \log(\sin x \cos x) \, dx $

$2I = \int_{0}^{\pi/2} \log \frac {2\sin x \cos x}{2} \, dx $

By properties of logarithms

$\log a – \log b = \log \frac {a}{b}$

So , we have

$2I = \int_{0}^{\pi/2} \log (2\sin x \cos x) – \log 2 \, dx $

$2I = \int_{0}^{\pi/2} \log (\sin 2x ) – \log 2 \, dx $

$2I = \int_{0}^{\pi/2} \log (\sin 2x) \, dx – \int_{0}^{\pi/2} \log 2 \, dx $ -(3)

Now lets calculate the Integral $\int_{0}^{\pi/2} \log (\sin 2x) \, dx $ separately

let 2x= t

$2 dx = dt$ or $ dx = \frac {dt}{2}$

Therefore

$\int_{0}^{\pi/2} \log (\sin 2x) \, dx = \frac {1}{2} \int_{0}^{\pi} \log (\sin t) \, dt $ -(4)

Now we know from properties of Definite integral

$\int_{0}^{2a} f(x) dx=2\int_{0}^{a} f(x)) dx$ if f(x) = f(2a-x)

here

$\log (\sin t) = \log (\sin (\pi -t)$

Hence

$\int_{0}^{\pi} \log (\sin t) \, dt = 2 \int_{0}^{\pi/2} \log (\sin t) \, dt$

putting this in (4), we get

$\int_{0}^{\pi/2} \log (\sin 2x) \, dx = \int_{0}^{\pi/2} \log (\sin t) \, dt $

Now $\int_{0}^{\pi/2} \log (\sin t) \, dt=\int_{0}^{\pi/2} \log(\sin x) \, dx =\int_{0}^{\pi/2} \log(\sin \pi/2 – x) \, dx= \int_{0}^{\pi/2} \log(\cos x \, dx=I $

So putting in (3) , we get

$2I= I – \int_{0}^{\pi/2} \log 2 \, dx $

$I =-\frac {\pi}{2} \log 2 $

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