Question

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

Answer 1

Hint:Here, use properties of definite integrals to convert sin x to cos x and add the two integrals. Using properties of logarithm, simplify the integral. Also using properties of trigonometric functions simplify the integral to get the result

Complete step-by-step answer:
Let $I = \int_0^{\dfrac{\pi }{2}} {\ln (\sin x)dx} $ …(i)
$ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\ln \left[ {\sin \left( {\dfrac{\pi }{2} - x} \right)} \right]} dx$ [Using property of definite integrals]
[Property of definite integral: In a definite integral, $\int_0^a {f(x)dx} = \int_0^a {f(x - a)dx} $]
$ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\ln (\cos x)dx} $ …(ii) [since, sin $\left( {\dfrac{\pi }{2} - x} \right)$ = cos x]
Adding equations (i) and (ii), we get
$2I = \int_0^{\dfrac{\pi }{2}} {\ln (\sin x)dx + } \int_0^{\dfrac{\pi }{2}} {\ln (\cos x)dx} $
$ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\left[ {\ln (\sin x) + \ln (\cos x)} \right]} dx$ [Using property of integrals]
[Property of definite integrals: $\int_0^{\dfrac{\pi }{2}} {f(x)dx} + \int_0^{\dfrac{\pi }{2}} {g(x)dx} = \int_0^{\dfrac{\pi }{2}} {\left[ {f(x) + g(x)} \right]dx} $]
$ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\ln (\sin x \times \cos x)dx} $
[Property of logarithm: ln (a) +ln (b) = ln (ab)]
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\ln \left[ {\dfrac{{2\sin x \times \cos x}}{2}} \right]dx} \]
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\ln \left[ {\dfrac{{\sin 2x}}{2}} \right]dx} \] [Since, 2 × sinx × cosx = sin 2x]
[Applying logarithm property: $\log \left( {\dfrac{a}{b}} \right) = \log a - \log b$]
$ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\left[ {\ln (\sin 2x) - \ln 2} \right]dx} $
$ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\ln \sin 2x} dx + \int_0^{\dfrac{\pi }{2}} {\ln 2} dx$ …(iii)
Assume, ${I_1} = \int_0^{\dfrac{\pi }{2}} {\ln \sin 2x} dx$
Putting 2x = y
For x = 0, y = 0
For $x = \dfrac{\pi }{2},y = \pi $
Differentiating both sides of (2x = y) with respect to x, we get
$2 = \dfrac{{dy}}{{dx}} \Rightarrow dx = \dfrac{{dy}}{2}$
Now, ${I_1} = \int_0^{\dfrac{\pi }{2}} {\ln (\sin y)\dfrac{{dy}}{2}} = \dfrac{1}{2}\int_0^{\dfrac{\pi }{2}} {\ln (\sin y)dy} $
${I_1} = \dfrac{1}{2}I$ [Since, $\int_a^b {f(x)dx} = \int_a^b {f(y)dy} $]
Putting value of ${I_1}$in equation (iii), we have
\[2I = \dfrac{I}{2} + \ln 2\mathop {\left[ x \right]}\nolimits_0^{\dfrac{\pi }{2}} \] $\left[ {\because \int_0^{\dfrac{\pi }{2}} {\ln 2dx = \ln 2\mathop {\left[ x \right]}\nolimits_0^{\dfrac{\pi }{2}} } } \right]$
$ \Rightarrow 2I - \dfrac{I}{2} = \dfrac{\pi }{2}\ln 2$
Separating I and constant parts
$ \Rightarrow \dfrac{{3I}}{2} = \dfrac{\pi }{2}\ln 2$
Multiplying both sides by 2
$ \Rightarrow 3I = \pi \ln 2$
$ \Rightarrow I = \dfrac{\pi }{3}\ln 2$

Note:In these types of questions, always use properties of definite integral do not go for actual integration steps. Try to use trigonometric identities, properties of definite integrals and logarithmic properties to simplify the given problem. Always careful while doing the steps and using properties. If we solve these types of integral problems directly it becomes much more complicated, and it may happen that you will not get the final results. These properties can be used only for definite integrals not for indefinite integrals. For doing these types of problems brush up trigonometric identities, and should also be aware about logarithmic basic properties, without which you will not be able to solve the problem.