Question

If we have the function $f(x)=\dfrac{2-x\cos x}{2+x\cos x}$ and $g(x)={{\log }_{e}}x$(x > 0) then the value of integral $\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{g(f(x))dx}$ is
(a) ${{\log }_{e}}3$
(b) ${{\log }_{e}}2$
(c) ${{\log }_{e}}e$
(d) ${{\log }_{e}}1$

Answer 1

Hint: To solve this question, firstly we will find the value of composite function, $g(f(x))$ where $f(x)=\dfrac{2-x\cos x}{2+x\cos x}$ and $g(x)={{\log }_{e}}x$. Then, we will substitute the value of $g(f(x))$ in integral $\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{g(f(x))dx}$ and let the integral be equals to I. after that we will replace x by –x and add the both integral. After that, we will simplify the integral and using properties of log we will obtain the value of integral.

Complete step-by-step solution:
Now let us find $g(f(x))$.
We are given that, $f(x)=\dfrac{2-x\cos x}{2+x\cos x}$ and $g(x)={{\log }_{e}}x$
The, we can say that $g\left( \left( \dfrac{2-x\cos x}{2+x\cos x} \right) \right)$
So, $g(f(x))={{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right)$
So, in question we are asked to evaluate the value of integral $\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{g(f(x))dx}$.
So, putting the value of $g(f(x))={{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right)$ in the integral $\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{g(f(x))dx}$, we get
$\Rightarrow \int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right)dx}$
So, let integral be equals to I
So, $I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right)dx}$…………... ( i )
Let us replace, x by – x, we get
$I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2-(-x)\cos x}{2+(-x)\cos x} \right)dx}$
On simplifying, we get
$I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2+x\cos x}{2-x\cos x} \right)dx}$………………... ( ii )
Adding ( i ) and ( ii ), we get
$I+I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right)dx}+\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2+x\cos x}{2-x\cos x} \right)dx}$
$\Rightarrow 2I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right)dx}+\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2+x\cos x}{2-x\cos x} \right)dx}$
As limits of both integral are same, so we can add both functions too,
So, we get
$2I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{\left( {{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right)+{{\log }_{e}}\left( \dfrac{2+x\cos x}{2-x\cos x} \right) \right)dx}$
On simplifying, we get
$2I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right)\left( \dfrac{2+x\cos x}{2-x\cos x} \right)dx}$, as we know that ${{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}xy$ .
On simplifying, we get
$2I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}1dx}$
As, ${{\log }_{e}}1$ is constant value, so we can pull this out of integral.
So, we get
$2I={{\log }_{e}}1\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{1dx}$
We know that, $\int{1.dx=x}$
So, $2I={{\log }_{e}}1.\{x\}_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}$
On putting, upper limit as $\dfrac{\pi }{4}$ and lower limit as $-\dfrac{\pi }{4}$, we get
\[2I={{\log }_{e}}1.\left( \dfrac{\pi }{4}-\left( -\dfrac{\pi }{4} \right) \right)\]
On simplifying, we get
\[2I={{\log }_{e}}1.\left( \dfrac{2\pi }{4} \right)\]
\[\Rightarrow I={{\log }_{e}}1.\left( \dfrac{2\pi }{2.4} \right)\]
Now, we know that \[{{\log }_{e}}1=0\], so we get
\[I=0\]
Hence, option ( d ) is correct.

Note: To solve such questions, one must know how we calculate the composite function when we are given two functions because we cannot proceed without this step. Also, remember the property of definite integration which is \[\int\limits_{a}^{b}{f(x)dx=F(a)-F(b)}\] ,where F is the integration of f ( x ) and a is lower limit and b is upper limit. One must know the properties and values of logarithmic function such as ${{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}xy$ and \[{{\log }_{e}}1=0\]. Try not to make any calculation error, while solving the integral.