Solve the integration \[\int{\cos x\log \left( \cos x \right)dx}=\sin x\log \left( \cos x \right)+\log \left| \sec x+\tan x \right|+f\left( x \right)+c\]
, then \[f\left( x \right)=?\]
(a) \[-\sin x\]
(b) \[\cos x\]
(c) \[\tan x\]
(d) \[\cot x\]

Hint: Apply integration by parts using the ILATE rule by assuming \[\log \left( \cos x \right)\] as the first function and \[\cos x\] as the second function. Simplify the obtained integral and apply integration by parts one more time by assuming \[\tan x\] as the first function and \[\sin x\] as the second function. Use the conversion, \[\sin x\] to simplify the expression obtained. Compare the obtained L.H.S with the given R.H.S and find the value of \[f\left( x \right)\].

Complete step-by-step solution
Here, we have provided with the expression: -
\[\Rightarrow \int{\cos x\log \left( \cos x \right)dx}=\sin x\log \left( \cos x \right)+\log \left| \sec x+\tan x \right|+f\left( x \right)+c\] and we have to find the value of \[f\left( x \right)\].
Now, here we can see that we have a product of two functions inside the integral sign, i. e, \[\cos x\] and \[\log \left( \cos x \right)\]. So, here we must apply the integration by parts method to find the integral. Here, we assume the given functions as function \[{{f}_{1}}\left( x \right)\] and function \[{{f}_{2}}\left( x \right)\] according to ILATE rule. ILATE stands for: -
I \[\to \] Inverse trigonometric functions
L \[\to \] Logarithmic functions
A \[\to \] Algebraic functions
T \[\to \] Trigonometric functions
E \[\to \] Exponential functions
In the ILATE rule we assume \[{{f}_{1}}\left( x \right)\] as the function which comes first according to the above given list. For example: - in the above question we have a trigonometric function \[\cos x\] and a logarithmic function \[\log \left( \cos x \right)\]. So, we will assume \[{{f}_{1}}\left( x \right)=\log \left( \cos x \right)\] and \[{{f}_{2}}\left( x \right)=\cos x\]. Now, the formula for the integral is given as: - \[\int{{{f}_{1}}\left( x \right).{{f}_{2}}\left( x \right)}={{f}_{1}}\left( x \right).\int{{{f}_{2}}\left( x \right)dx}-\int{\left( \int{{{f}_{2}}\left( x \right)dx} \right).f_{1}^{'}\left( x \right)dx}\].
Here, \[f_{1}^{'}\left( x \right)=\dfrac{d\left[ {{f}_{1}}\left( x \right) \right]}{dx}\].
Now, using the ILATE rule we have,
\[\Rightarrow \] L.H.S = \[\int{\cos x\log \left( \cos x \right)dx}\]
\[\Rightarrow \] L.H.S = \[\log \left( \cos x \right)\int{\cos xdx}-\int{\left\{ \left( \int{\cos xdx} \right).\dfrac{d\left[ \log \left( \cos x \right) \right]}{dx} \right\}dx}\]
\[\Rightarrow \] L.H.S = \[\log \left( \cos x \right).\sin x-\int{\sin x\times \dfrac{1}{\cos x}\times \left( -\sin x \right)dx}\]
\[\Rightarrow \] L.H.S = \[\sin x\log \left( \cos x \right)+\int{\left( \dfrac{\sin x}{\cos x} \right)\times \sin xdx}\]
Using the conversion, \[\dfrac{\sin x}{\cos x}=\tan x\], we get,
\[\Rightarrow \] L.H.S = \[\sin x\log \left( \cos x \right)+\int{\tan x\times \sin xdx}\]
Now, in the above relation we have a product of \[\tan x\] and \[\sin x\] and both of them are trigonometric functions. Here, we know the integration of \[\sin x\] can be found easily, so assuming \[{{f}_{1}}\left( x \right)=\tan x\] and \[{{f}_{2}}\left( x \right)=\sin x\], we get,
\[\Rightarrow \] L.H.S = \[\sin x\log \left( \cos x \right)+\tan x\int{\sin xdx}-\int{\left\{ \left( \int{\sin xdx} \right).\dfrac{d\left[ \tan x \right]}{dx} \right\}dx}\]
\[\Rightarrow \] L.H.S = \[\sin x\log \left( \cos x \right)+\tan x.\left( -\cos x \right)-\int{\left( -\cos x \right).{{\sec }^{2}}xdx}\]
\[\Rightarrow \] L.H.S = \[\sin x\log \left( \cos x \right)-\cos x.\tan x+\int{\left( \cos x.\sec x \right).\sec xdx}\]
Using the identity, \[\tan x=\dfrac{\sin x}{\cos x}\] and \[\sec x.\cos x=1\], we get,
\[\Rightarrow \] L.H.S = \[\sin x\log \left( \cos x \right)-\cos x.\dfrac{\sin x}{\cos x}+\int{1.\sec xdx}\]
\[\Rightarrow \] L.H.S = \[\sin x\log \left( \cos x \right)-\sin x+\int{\sec xdx}\]
We know that \[\int{\sec xdx}=\log \left| \sec x+\tan x \right|+c\], where ‘c’ is the constant of integration.
Now, comparing the obtained L.H.S with the given R.H.S we can clearly conclude that, \[f\left( x \right)=-\sin x\].
Hence, option (a) is the correct answer.

Note: One may note that in the product of \[\tan x\] and \[\sin x\] it was confusing to choose the first function because both were trigonometric functions and the ILATE rule could not be applied. Here, we saw that we can easily find the integration of \[\sin x\] and that is why it was assumed to be the second function. You must remember the ILATE rule and basic integration and differentiation formulas to solve the question.
.