Question

Find the integral \[\int\limits_{-{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\log \left( \dfrac{2-\sin \theta }{2+\sin \theta } \right)}d\theta =\]
A. \[0\]
B. \[1\]
C. \[2\]
D. None of these

Answer 1

Hint: In this question, we are to find the given integral. The given integral is in the interval $[-1,1]$. So, we can go for an even or odd function integral. By applying $\theta =-\theta $ in the given function, we can find that the function is an even or odd function. According to the type of the function, we can evaluate the integral.



Formula Used:Definite integral:
Consider a function $f(x)$ is defined on $[a,b]$. If the integral of this function, $\int{f(x)dx=F(x)}$, then $F(b)-F(a)$ is called the definite integral of the function $f(x)$ over $[a,b]$.
I.e.,$\int\limits_{a}^{b}{f(x)dx}=\left. F(x) \right|_{a}^{b}=F(b)-F(a)$
Here $a$ (lower limit) and $b$(upper limit).
\[\int\limits_{a}^{b}{f(x)dx}=\int\limits_{a}^{b}{f(t)dt}\]
Some of the properties of the definite integrals are:
1) Interchanging the limits: \[\int\limits_{a}^{b}{f(x)dx=-}\int\limits_{b}^{a}{f(x)dx}\]
2) If $a3) $\int\limits_{0}^{a}{f(x)dx}=\int\limits_{0}^{a}{f(a-x)dx}$
4) $\int\limits_{-a}^{a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx}$ if $f(x)$ is an even function
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
5) $\begin{align}
  & \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
 & \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$
6) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\sin x)}{f(\sin x)+f(\cos x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cos x)}{f(\sin x)+f(\cos x)}dx=\dfrac{\pi }{4}}}$
7) $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\tan x)}{f(\tan x)+f(\cot x)}dx=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{f(\cot x)}{f(\tan x)+f(\cot x)}dx=\dfrac{\pi }{4}}}$



Complete step by step solution:Given integral is
\[I=\int\limits_{-{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\log \left( \dfrac{2-\sin \theta }{2+\sin \theta } \right)}d\theta \]
Since the limits of the given function are in the form of $[-a, a]$, we need to find the type of the function.
Consider the function in the given integral as
\[f(\theta )=\log \left( \dfrac{2-\sin \theta }{2+\sin \theta } \right)\]
So, substituting $\theta =-\theta $, we get
\[\begin{align}
  & f(-\theta )=\log \left( \dfrac{2-\sin (-\theta )}{2+\sin (-\theta )} \right) \\
 & \text{ }=\log \left( \dfrac{2+\sin \theta }{2-\sin \theta } \right) \\
 & \text{ }=\log (2+\sin \theta )-\log (2-\sin \theta ) \\
 & \text{ }=-\left[ \log (2-\sin \theta )-\log (2+\sin \theta ) \right] \\
 & \text{ }=-\log \left( \dfrac{2-\sin \theta }{2+\sin \theta } \right) \\
 & \text{ }=-f(\theta ) \\
\end{align}\]
Since \[f(-\theta )=-f(\theta )\], the function in the given integral is an odd function. Thus, according to the formula we have
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
The given integral becomes
\[I=\int\limits_{-{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\log \left( \dfrac{2-\sin \theta }{2+\sin \theta } \right)}d\theta =0\]



Option ‘A’ is correct



Note: In this question, we have trigonometric function in the integral. So, it is a little difficult to solve by a normal ILET integration method. Since the given integral has the limits in the interval of $[-a,a]$ type, we can use the predefined formulae, where the type of the function plays the role. So, in order to know the type of the function, we need to substitute $\theta =-\theta $ in the function $f(\theta )$. Then, according to the type of the function i.e., either it is an even or odd function, we can evaluate the integral of the given function within the interval.