Question

Find the integration:- $\int_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log \sin 2x} \right)} dx$

Answer 1

Here we need to find the value of the given definite integral. For that, we will first simplify the terms inside the bracket. We will first use the basic trigonometric identities and then we will use the properties of the logarithmic function. We will also use the basic properties of definite integrals to simplify it. From there, we will get the simplified term, and we will then integrate it to get the final solution.

Complete step-by-step answer:
We need to find the value of $\int_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log \sin 2x} \right)} dx$
Let $I$ be the required definite integration.
$\Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log \sin 2x} \right)} dx$ ………. $\left( 1 \right)$
We know from the trigonometric identities that $\sin 2x = 2 \cdot \sin x \cdot \cos x$.
Using this identity in equation $\left( 1 \right)$, we get
$\Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log 2 \cdot \sin x \cdot \cos x} \right)} dx$
We know the property of logarithm that $\log ab = \log a + \log b$ .
We will use this property here, we get
$\Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log 2 - \log \sin x - \log \cos x} \right)} dx$
On further simplifying the terms present inside the bracket, we get
$\Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\left( {\log \sin x - \log 2 - \log \cos x} \right)} dx$
We can write this equation as
$\Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\log \sin x.dx - \int_0^{\dfrac{\pi }{2}} {\log 2.dx} - \int_0^{\dfrac{\pi }{2}} {\log \cos x} } .dx$
We can one property of definite integral that $\int_0^a {f\left( x \right)dx = \int_0^a {f\left( {a - x} \right)dx} }$
Using this property for the third term, we get
$\Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\log \sin x.dx - \int_0^{\dfrac{\pi }{2}} {\log 2.dx} - \int_0^{\dfrac{\pi }{2}} {\log \cos \left( {\dfrac{\pi }{2} - x} \right)} } .dx$
We know from periodic identities that $\cos \left( {\dfrac{\pi }{2} - x} \right) = \sin x$.
Using this identity for the third term, we get
$\Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\log \sin x - \int_0^{\dfrac{\pi }{2}} {\log 2} - \int_0^{\dfrac{\pi }{2}} {\log \sin x} }$
On further simplification, we get
$\Rightarrow I = - \int_0^{\dfrac{\pi }{2}} {\log 2} .dx$
Taking constant out of the integration, we get
$\Rightarrow I = - \log 2\int_0^{\dfrac{\pi }{2}} 1 .dx$
On integrating the term, we get
$\Rightarrow I = - \log 2\left[ x \right]_0^{\dfrac{\pi }{2}}$
On simplifying the term, we get
$\Rightarrow I = - \log 2\left( {\dfrac{\pi }{2} - 0} \right)$
On further simplification, we get
$\Rightarrow I = - \dfrac{\pi }{2}\log 2$
Hence, the value of $\int_0^{\dfrac{\pi }{2}} {\left( {2\log \sin x - \log \sin 2x} \right)} dx$ is equal to $- \dfrac{\pi }{2}\log 2$.

Note: Here, we need to be careful while integrating and therefore we need to know the basics of integration. Integration is a method by which we can find summation of discrete data.
While solving this question we might make a mistake by leaving the answer at $I = - \log 2\left[ x \right]_0^{\dfrac{\pi }{2}}$. This will be wrong because this is a question of definite integral and not an infinite integral. So, we need to substitute the limits to find the required answer.