Question

Evaluate $\int\limits_{{}^{-1}/{}_{2}}^{{}^{1}/{}_{2}}{(\cos x)\left[ \log \left( \dfrac{1-x}{1+x} \right) \right]}dx=$
A. $0$
B. $1$
C. ${{e}^{{}^{1}/{}_{2}}}$
D. $2{{e}^{{}^{1}/{}_{2}}}$

Answer 1

Hint: In this question, we are to find the given integral. A definite integral is easily evaluated by using its properties. One such property applied for the given integral is even and odd function integral. Depending on the type of the function, the integral is evaluated.



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 as 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) $\begin{align}
  & \int\limits_{0}^{2a}{f(x)dx}=2\int\limits_{0}^{a}{f(x)dx};\text{if }f(2a-x)=f(x) \\
 & \text{ }=0\text{ if }f(2a-x)=-f(x) \\
\end{align}$

Complete step by step solution:Given integral is
$I=\int\limits_{{}^{-1}/{}_{2}}^{{}^{1}/{}_{2}}{(\cos x)\left[ \log \left( \dfrac{1-x}{1+x} \right) \right]}dx$
Consider the given function as
$f(x)=(\cos x)\left[ \log \left( \dfrac{1-x}{1+x} \right) \right]$
To evaluate the given integral, we need to know the type of the function.
So, substituting $x=-x$ in the given function $f(x)$
Then,
$\begin{align}
  & f(-x)=(\cos (-x))\left[ \log \left( \dfrac{1-(-x)}{1+(-x)} \right) \right] \\
 & \text{ }=(\cos x)\left[ \log \left( \dfrac{1+x}{1-x} \right) \right] \\
 & \text{ }=(\cos x)\left[ \log (1+x)-\log (1-x) \right] \\
 & \text{ }=-(\cos x)\left[ \log (1-x)-\log (1+x) \right] \\
 & \text{ }=-(\cos x)\left[ \log \left( \dfrac{1-x}{1+x} \right) \right] \\
 & \text{ }=-f(x) \\
\end{align}$
Since $f(x)=-f(x)$, the function in the given integral is an odd function.
So, by the property, we have
$\int\limits_{-a}^{a}{f(x)dx}=0$ if $f(x)$ is an odd function
Therefore,
$I=\int\limits_{{}^{-1}/{}_{2}}^{{}^{1}/{}_{2}}{(\cos x)\left[ \log \left( \dfrac{1-x}{1+x} \right) \right]}dx=0$



Option ‘A’ is correct



Note: Here we need to remember that, to apply this property, first we need to check the interval of the integral. If the interval is in the form of $[-a,a]$, then the function in the integral is to be verified for the type of the function by substituting $x=-x$. Thus, depending on the type of function, the integration is to be evaluated.