Question

Value of $\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\cos 4x \right)dx}$ is
(a) $\dfrac{\pi }{4}\ln 2$
(b) $-\dfrac{\pi }{4}\ln 2$
(c) $-\dfrac{\pi }{2}\ln 2$
(d) 0

Answer 1

This is a problem of integration where we have to find the value of a given definite integral. So, we will start with taking the whole integration as I and will use different integration properties to get an easier form of the problem. Then logarithmic properties will be used to get ahead with the problem and get our solution.

Complete step by step solution:
According to the problem, we are to find the value of $\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\cos 4x \right)dx}$
Now, we know from the integration identities, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$
So, let us consider, I = $\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\cos 4x \right)dx}$……..[1]
Now, using this property, $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\cos 4\left( \dfrac{\pi }{4}-x \right) \right)dx}$
Simplifying, $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\cos \left( \pi -4x \right) \right)dx}$
Again, we know, $\cos \left( \pi -x \right)=-\cos x$ , as it is in the 2nd quadrant.
So, we are getting, $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1-\cos \left( 4x \right) \right)dx}$…….[2]
Now, adding 1 and 2, we get,
$2I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\cos \left( 4x \right) \right)dx}+\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1-\cos \left( 4x \right) \right)dx}$
Writing them together,
$2I=\int\limits_{0}^{\dfrac{\pi }{4}}{\left[ \ln \left( 1+\cos \left( 4x \right) \right)dx+\ln \left( 1-\cos \left( 4x \right) \right)dx \right]}$
And, $\ln m+\ln p=\ln mp$ as the logarithmic formulas suggest.
Hence, we have,
$2I=\int\limits_{0}^{\dfrac{\pi }{4}}{\left[ \ln \left( 1+\cos \left( 4x \right) \right)\left( 1-\cos \left( 4x \right) \right)dx \right]}$
Simplifying,
$2I=\int\limits_{0}^{\dfrac{\pi }{4}}{\left[ \ln \left( 1-{{\cos }^{2}}\left( 4x \right) \right)dx \right]}$
Again, $1-{{\cos }^{2}}x={{\sin }^{2}}x$ gives us,
$2I=\int\limits_{0}^{\dfrac{\pi }{4}}{\left[ \ln \left( {{\sin }^{2}}\left( 4x \right) \right)dx \right]}$
Using, $\log {{a}^{2}}=2\log a$ we get,
$2I=\int\limits_{0}^{\dfrac{\pi }{4}}{2\ln \left( \sin \left( 4x \right) \right)dx}=2\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( \sin \left( 4x \right) \right)dx}$
Cancelling 2 from both sides,
$I=\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( \sin \left( 4x \right) \right)dx}$
Again,$\int\limits_{0}^{2a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( a-x \right)dx}$ if $f\left( a-x \right)=f\left( x \right)$ .
Here, for $f\left( x \right)=\ln \left( \sin 4x \right)$ and $f\left( \dfrac{\pi }{4}-x \right)=\ln \left( \sin 4\left( \dfrac{\pi }{4}-x \right) \right)=\ln \left( \sin \left( \pi -4x \right) \right)=\ln \left( \sin 4x \right)$
So, $I=2\int\limits_{0}^{\dfrac{\pi }{8}}{\ln \left( \sin \left( 4x \right) \right)dx}$…….[3]
Again, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$
Thus, $I=2\int\limits_{0}^{\dfrac{\pi }{8}}{\ln \left( \sin \left( 4\left( \dfrac{\pi }{8}-x \right) \right) \right)dx}$
$\Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{8}}{\ln \left( \sin \left( \dfrac{\pi }{2}-4x \right) \right)dx}=2\int\limits_{0}^{\dfrac{\pi }{8}}{\ln \left( \cos 4x \right)dx}$…..[4]
Adding 4 and 5,
$2I=2\int\limits_{0}^{\dfrac{\pi }{8}}{\left( \ln \left( \sin \left( 4x \right) \right)+\ln \left( \cos \left( 4x \right) \right) \right)dx}$
Similarly, simplifying,
$I=\int\limits_{0}^{\dfrac{\pi }{8}}{\left( \ln \left( \sin \left( 4x \right).\cos \left( 4x \right) \right) \right)dx}$
Again,
$I=\int\limits_{0}^{\dfrac{\pi }{8}}{\left( \ln \left( \dfrac{2\sin \left( 4x \right).\cos \left( 4x \right)}{2} \right) \right)dx}$
Using trigonometric identities,
$I=\int\limits_{0}^{\dfrac{\pi }{8}}{\left( \ln \left( \dfrac{\sin \left( 8x \right)}{2} \right) \right)dx}$
Now, again using the logarithmic identities, $\log a-\log b=\log \dfrac{a}{b}$ ,
We are getting, $I=\int\limits_{0}^{\dfrac{\pi }{8}}{\left( \ln \left( \sin 8x \right)-\ln 2 \right)dx}$
Writing them differently,
$I=\int\limits_{0}^{\dfrac{\pi }{8}}{\ln \left( \sin 8x \right)dx}-\int\limits_{0}^{\dfrac{\pi }{8}}{\ln 2dx}$
Let us consider, ${{I}_{1}}=\int\limits_{0}^{\dfrac{\pi }{8}}{\ln \left( \sin 8x \right)dx}$
So, to integrate this we are using the substitution $8x=t$ , $\Rightarrow dx=\dfrac{1}{8}dt$
And the limits will be,
When $x=0,t=0$ and $x=\dfrac{\pi }{8},t=\pi$
${{I}_{1}}=\int\limits_{0}^{\pi }{\ln \left( \sin t \right)\dfrac{dt}{8}}$
Simplifying,
${{I}_{1}}=\dfrac{1}{8}\int\limits_{0}^{\pi }{\ln \left( \sin t \right)dt}=\dfrac{I}{2}$
Hence, we have,
$I={{I}_{1}}-\int\limits_{0}^{\dfrac{\pi }{8}}{\ln 2dx}$
$\Rightarrow I=\dfrac{I}{2}-\int\limits_{0}^{\dfrac{\pi }{8}}{\ln 2dx}$
So,
$\Rightarrow \dfrac{I}{2}=-\ln 2\left[ \dfrac{\pi }{8}-0 \right]$
Thus, we can conclude,
$\Rightarrow I=-\dfrac{\pi }{4}\ln 2$
So, $\int\limits_{0}^{\dfrac{\pi }{4}}{\ln \left( 1+\cos 4x \right)dx}=-\dfrac{\pi }{4}\ln 2$

So, the correct answer is “Option (b)”.

Note: This given problem is a very complicated one and to be dealt with very carefully. Some identities of integration like $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$, is to be analyzed properly and then to be used. If the given function doesn’t satisfy the given conditions we can conclude that the property can not be used. Similarly for putting the value of limits and integration, we need to be extra cautious and find the proper solution.