Question

Integrate $\log (\sin x)$ from $0$ to $\dfrac{\pi }{2}$.

Answer 1

Hint: To solve the given definite integral, we will use some properties of definite integral along with some other properties of logarithm and properties of integration. The properties of definite integral used while computing the definite integral are:
$\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)} } dx$,
$\int\limits_0^{2a} {f(x)dx = 2\int\limits_0^a {f(x)dx,{\text{ if }}f(2a - x) = f(x)} } $
$\int\limits_a^b {f(x)} dx = \int\limits_a^b {f(t)dt} $
These properties are used in solving the integration in the same order along with other properties and methods of solving integration.

Complete step by step answer:
Let us assume the given definite integral as,
${I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x)dx} - - - - (1)$
Now, using the definite integral property, $\int\limits_0^a {f(x)dx = \int\limits_0^a {f(a - x)} } dx$, we get,
$ \Rightarrow {I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \left[ {\sin \left( {\dfrac{\pi }{2} - x} \right)} \right]dx} $
Now, we know the trigonometric formula $\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x$. So, we get,
$ \Rightarrow {I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx} - - - - \left( 2 \right)$
Now, adding equations $(1)$ and $(2)$, we get,
$ \Rightarrow {I_1} + {I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x)dx} + \int\limits_0^{\dfrac{\pi }{2}} {\log (\cos x)dx} $
Now, we know the property of logarithm $\log a + \log b = \log \left( {ab} \right)$. So, using this property in the expression, we have,
$ \Rightarrow 2{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x \times \cos x)dx} $

Now, multiplying and dividing $2$ with$\sin x\cos x$,
$ \Rightarrow 2{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log \left[ {\dfrac{{2\sin x\cos x}}{2}} \right]dx} $
Using the property of logarithm $\log a - \log b = \log \left( {\dfrac{a}{b}} \right)$,
$ \Rightarrow 2{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log (2\sin x\cos x) - \log 2} \right]dx} $
Now, we know the trigonometric double angle formula for sine as $\sin 2x = 2\sin x\cos x$.
$ \Rightarrow 2{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\left[ {\log (\sin 2x) - \log 2} \right]dx} $
Separating both the integrals, we get,
$ \Rightarrow 2{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin 2x)dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} - - - - (3)$
Let, ${I_2} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin 2x)dx} $
Now, let $2x = t$.
Differentiating both sides with respect to x, we get,
$ \Rightarrow 2 = \dfrac{{dt}}{{dx}}$
$ \Rightarrow dx = \dfrac{{dt}}{2}$

Now, for adjusting the lower limit and upper limit of the integral, we have,

$x$	$t = 2x$
$0$	$2(0) = 0$
$\dfrac{\pi }{2}$	$2\left( {\dfrac{\pi }{2}} \right) = \pi $

Therefore, putting the values of t and dt. Also, changing the limits, we get,
${I_2} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin 2x)dx} $
$ \Rightarrow {I_2} = \int\limits_0^\pi {\log (\sin t)\dfrac{{dt}}{2}} $
$ \Rightarrow {I_2} = \dfrac{1}{2}\int\limits_0^\pi {\log (\sin t)dt} $
Now, using the property, $\int\limits_0^{2a} {f(x)dx = 2\int\limits_0^a {f(x)dx,{\text{ if }}f(2a - x) = f(x)} } $,
Here, let $f(t) = \log \sin t$.
Then, $f(2a - t) = f(2\pi - t) = \log [\sin (2\pi - t)] = \log (\sin t)$
We get, $f(t) = f(2a - t)$
$ \Rightarrow {I_2} = \dfrac{1}{2}\int\limits_0^\pi {\log (\sin t)dt} $
$ \Rightarrow {I_2} = \dfrac{1}{2} \times 2\int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t)dt} $
$ \Rightarrow {I_2} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t)dt} $
Now, using the property, $\int\limits_a^b {f(x)} dx = \int\limits_a^b {f(t)dt} $, we get,
${I_2} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x)dx} $
Putting, the values of ${I_2}$in equation $(3)$, we get,
$2{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin 2x)dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} $
$ \Rightarrow 2{I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x)dx} - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} $
Now, we can see that ${I_2} = {I_1} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x)dx} $
Therefore, we can write,
$ \Rightarrow 2{I_1} = {I_1} - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} $
Now, subtracting ${I_1}$from both sides, we get,
$ \Rightarrow {I_1} = - \int\limits_0^{\dfrac{\pi }{2}} {\log 2dx} $
Taking constant term outside of the integral, we get,
$ \Rightarrow {I_1} = - \log 2\int\limits_0^{\dfrac{\pi }{2}} {dx} $
Now, we know that the derivative of $1$ with respect to x is x. So, putting in the limits of the integral, we get,
$ \Rightarrow {I_1} = - \log 2\left[ {\dfrac{\pi }{2} - 0} \right]$
$ \Rightarrow {I_1} = - \log 2\left[ {\dfrac{\pi }{2}} \right]$
$ \Rightarrow {I_1} = - \dfrac{\pi }{2}\log 2$
$ \therefore \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x)dx} = - \dfrac{\pi }{2}\log 2$
Note:The given problem requires thorough knowledge of the properties involving definite integrals. One should not forget to change the limits of integral while changing the variable in the integration. We should remember the integrals of some basic functions in order to tackle such questions. We must take care of the calculations while solving the problem.