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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) the...

If we have the function f(x)=2−xcos⁡x2+xcos⁡xf(x)=\dfrac{2-x\cos x}{2+x\cos x} and g(x)=log⁡exg(x)={{\log }_{e}}x(x > 0) then the value of integral ∫−π4π4g(f(x))dx\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{g(f(x))dx} is
(a) log⁡e3{{\log }_{e}}3
(b) log⁡e2{{\log }_{e}}2
(c) log⁡ee{{\log }_{e}}e
(d) log⁡e1{{\log }_{e}}1

Explanation

Solution

To solve this question, firstly we will find the value of composite function, g(f(x))g(f(x)) where f(x)=2−xcos⁡x2+xcos⁡xf(x)=\dfrac{2-x\cos x}{2+x\cos x} and g(x)=log⁡exg(x)={{\log }_{e}}x. Then, we will substitute the value of g(f(x))g(f(x)) in integral ∫−π4π4g(f(x))dx\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))g(f(x)).
We are given that, f(x)=2−xcos⁡x2+xcos⁡xf(x)=\dfrac{2-x\cos x}{2+x\cos x} and g(x)=log⁡exg(x)={{\log }_{e}}x
The, we can say that g((2−xcos⁡x2+xcos⁡x))g\left( \left( \dfrac{2-x\cos x}{2+x\cos x} \right) \right)
So, g(f(x))=log⁡e(2−xcos⁡x2+xcos⁡x)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 ∫−π4π4g(f(x))dx\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{g(f(x))dx}.
So, putting the value of g(f(x))=log⁡e(2−xcos⁡x2+xcos⁡x)g(f(x))={{\log }_{e}}\left( \dfrac{2-x\cos x}{2+x\cos x} \right) in the integral ∫−π4π4g(f(x))dx\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{g(f(x))dx}, we get
⇒∫−π4π4log⁡e(2−xcos⁡x2+xcos⁡x)dx\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=∫−π4π4log⁡e(2−xcos⁡x2+xcos⁡x)dxI=\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=∫−π4π4log⁡e(2−(−x)cos⁡x2+(−x)cos⁡x)dxI=\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=∫−π4π4log⁡e(2+xcos⁡x2−xcos⁡x)dxI=\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=∫−π4π4log⁡e(2−xcos⁡x2+xcos⁡x)dx+∫−π4π4log⁡e(2+xcos⁡x2−xcos⁡x)dxI+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}
⇒2I=∫−π4π4log⁡e(2−xcos⁡x2+xcos⁡x)dx+∫−π4π4log⁡e(2+xcos⁡x2−xcos⁡x)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=∫−π4π4(log⁡e(2−xcos⁡x2+xcos⁡x)+log⁡e(2+xcos⁡x2−xcos⁡x))dx2I=\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=∫−π4π4log⁡e(2−xcos⁡x2+xcos⁡x)(2+xcos⁡x2−xcos⁡x)dx2I=\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⁡ax+log⁡ay=log⁡axy{{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}xy .
On simplifying, we get
2I=∫−π4π4log⁡e1dx2I=\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{{{\log }_{e}}1dx}
As, log⁡e1{{\log }_{e}}1 is constant value, so we can pull this out of integral.
So, we get
2I=log⁡e1∫−π4π41dx2I={{\log }_{e}}1\int\limits_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}{1dx}
We know that, ∫1.dx=x\int{1.dx=x}
So, 2I=log⁡e1.x−π4π42I={{\log }_{e}}1.\\{x\\}_{-\dfrac{\pi }{4}}^{\dfrac{\pi }{4}}
On putting, upper limit as π4\dfrac{\pi }{4} and lower limit as −π4-\dfrac{\pi }{4}, we get
2I=log⁡e1.(π4−(−π4))2I={{\log }_{e}}1.\left( \dfrac{\pi }{4}-\left( -\dfrac{\pi }{4} \right) \right)
On simplifying, we get
2I=log⁡e1.(2π4)2I={{\log }_{e}}1.\left( \dfrac{2\pi }{4} \right)
⇒I=log⁡e1.(2π2.4)\Rightarrow I={{\log }_{e}}1.\left( \dfrac{2\pi }{2.4} \right)
Now, we know that log⁡e1=0{{\log }_{e}}1=0, so we get
I=0I=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 ∫abf(x)dx=F(a)−F(b)\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⁡ax+log⁡ay=log⁡axy{{\log }_{a}}x+{{\log }_{a}}y={{\log }_{a}}xy and log⁡e1=0{{\log }_{e}}1=0. Try not to make any calculation error, while solving the integral.