Question

Evaluate $\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx}$

Answer 1

Hint : As we can see that in the above question we have given the upper limit and the lower limit and then we have to integrate that. We can solve the above question with the help of logarithm formulas and the trigonometric identities. We know the basic trigonometric identity i.e. $\sin 2x = 2\sin x\cos x$ . In the logarithm if we have $\log a + \log b$ , we can write it as $\log ab$ .

Complete step-by-step answer :
Here we have $\int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx}$ . Let us assume
$I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx}$ . It is equation $(1)$ .
We can write the above as
$I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin \left( {\dfrac{\pi }{2} - x} \right)dx}$ , as we know that $\sin (90 - \theta ) = \cos \theta$ .
We have written here $\dfrac{\pi }{2} - x$ which means $\dfrac{{180}}{2} - x = 90 - x$ .
So by applying this we can write $I = \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx}$ , this is equation $(2)$ . We will add the equations one and two, so we have:
$I + I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} + \int\limits_0^{\dfrac{\pi }{2}} {\log \cos xdx}$ .
On further solving we have:
$2I = \int\limits_0^{\dfrac{\pi }{2}} {\log \sin xdx} + \log \cos xdx$ . By applying the above logarithm formula we can write : $2I = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x} \cos x)dx$. By multiplying and dividing the numerator and denominator by $2$ we have $\int\limits_0^{\dfrac{\pi }{2}} {\log (\dfrac{{2\sin x\cos x}}{2}} )dx$.
Again with the above trigonometric identity we can write it as $\int\limits_0^{\dfrac{\pi }{2}} {\log (\dfrac{{\sin 2x}}{2}} )dx$.
We can write it as $\int\limits_0^{\dfrac{\pi }{2}} {\log (\sin 2x} )dx - \int\limits_0^{\dfrac{\pi }{2}} {\log (2} )dx$. Now let us assume $2x = t$ .
Let us separate the left hand side $2I$ as ${I_1},{I_2}$ . So we can write
${I_2} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin 2x} )dx$.
Now we solve this. If we have $2x = t$ , after differentiating this with respect to $x$ we have $2 = \dfrac{{dt}}{{dx}} \Rightarrow dx = \dfrac{{dt}}{2}$ .
Now by putting the values of t and dt and changing the limits. We have
${I_2} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t} )\dfrac{{dt}}{2} \Rightarrow \dfrac{1}{2}\int\limits_0^\pi {\log (\sin t} )dt$.
We will use the property which says that
$\int\limits_0^{2a} {f(x)dx = 2\int\limits_0^a {f(x)dx,\,if\,f(2a - x) = f(x)} }$ .
On comparing from the formula we have $a = \pi$ . So here we have $f(t) = \log \sin t$ , by putting the value $f(2a - t) = f(2\pi - t) = \log \sin (2\pi - t) = \log \sin t$
We can write the above as $\dfrac{1}{2}\int\limits_0^\pi {\log (\sin t} )dt = \dfrac{1}{2} \times 2\int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t} )dt$. It gives us the value ${I_2} = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin t} )dt$. We will use another property now i.e. $\int\limits_a^b {f(x)dx = \int\limits_a^b {f(t)dt} }$ .
We will put the value of ${I_2}$ back in the equation we have $2I = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin 2x} )dx - \int\limits_0^{\dfrac{\pi }{2}} {\log (2\left( {dx} \right)}$.
On further solving we have $2I = \int\limits_0^{\dfrac{\pi }{2}} {\log (\sin x} )dx - \log (2)\int\limits_0^{\dfrac{\pi }{2}} {1.dx} \Rightarrow {I_1} - \log (2)(x)_0^{\dfrac{\pi }{2}}$.
By taking the similar terms to the left hand side and applying the value of limits we have $2I - I = - \log 2\left[ {\dfrac{\pi }{2} - 0} \right] \Rightarrow I = - \log 2\left[ {\dfrac{\pi }{2}} \right]$ .
We know the property that $\log {a^m} = m\log a$ . So by applying this we can write $I = \dfrac{{ - \pi }}{2}\log 2$ .
Hence the required answer is $I = \dfrac{{ - \pi }}{2}\log 2$ .
So, the correct answer is “ $I = \dfrac{{ - \pi }}{2}\log 2$ ”.

Note : We have used the rule of fraction of logarithm in the above solution. It says that $\log \dfrac{a}{b}$ can be written as $\log (a) - \log (b)$ . Before solving this kind of question we should always remember the rules of logarithms and the trigonometric identities and avoid the calculation mistakes. We have also used the integral property with upper and lower limits i.e. $\int\limits_0^{2a} {f(x)dx = 2\int\limits_0^a {f(x)dx,\,if\,f(2a - x) = f(x)} }$