Question

Evaluate the integral \[\int{\cos \left( {{\log }_{e}}x \right)dx}\] where \[c\] is the constant of integration.
(a) \[\dfrac{x}{2}\left( \cos \left( {{\log }_{e}}x \right)-\sin \left( {{\log }_{e}}x \right) \right)+c\]
(b) \[\dfrac{x}{2}\left( \cos \left( {{\log }_{e}}x \right)+\sin \left( {{\log }_{e}}x \right) \right)+c\]
(c) \[x\left( \cos \left( {{\log }_{e}}x \right)+\sin \left( {{\log }_{e}}x \right) \right)+c\]
(d) \[x\left( \cos \left( {{\log }_{e}}x \right)-\sin \left( {{\log }_{e}}x \right) \right)+c\]

Answer 1

Hint: In this question, in order to evaluate the definite integral \[\int{\cos \left( {{\log }_{e}}x \right)dx}\], we will first substitute the value \[{{\log }_{e}}x=t\] which implies that \[x={{e}^{t}}\] and then differentiate \[x={{e}^{t}}\] to get \[dx={{e}^{t}}dt\]. We will then substitute both \[{{\log }_{e}}x=t\] and \[dx={{e}^{t}}dt\] in the above integral to simplify it into \[\int{{{e}^{t}}\cos tdt}\]. Then we will use integration by parts in the integral \[\int{{{e}^{t}}\cos tdt}\]. We will then evaluate the same in order to get the desired answer.

Complete step by step answer:
Let \[I\] denote the integral \[\int{\cos \left( {{\log }_{e}}x \right)dx}\].
That is, let \[I=\int{\cos \left( {{\log }_{e}}x \right)dx}\].
Now on substituting the value \[{{\log }_{e}}x=t\] in the integrant of the above integral.
That is we have \[x={{e}^{t}}\]
On differentiating \[x={{e}^{t}}\], we will get
\[dx={{e}^{t}}dt\]
We will now substitute both the values \[{{\log }_{e}}x=t\] and \[dx={{e}^{t}}dt\] in the above integral \[I=\int{\cos \left( {{\log }_{e}}x \right)dx}\] to get
\[\begin{align}
  & I=\int{\cos \left( {{\log }_{e}}x \right)dx} \\
 & =\int{{{e}^{t}}\cos tdt}
\end{align}\]
Since we know that by integration by parts we have \[\int{ab=a\int{b-\int{{{a}^{'}}\int{b}}}}\]
by taking the first function and the second function usinf the rule of ILATE where
I stands for Integration of the function,
L stands for the logarithmic function,
A stands for algebraic expression,
T stands for trigonometric function and
E stands for exponential function.
Now using formula of integration by parts on both side of the above integral, we get
\[\begin{align}
  & I=\int{{{e}^{t}}\cos tdt} \\
 & =\cos t\int{{{e}^{t}}dt-\int{\dfrac{d}{dt}\left( \cos t \right)\int{{{e}^{t}}dt}dt}}................(1)
\end{align}\]
Now since we have \[\dfrac{d}{dt}\left( \cos t \right)=\sin t\] and \[\int{{{e}^{t}}dt}={{e}^{t}}+c\] where \[c\] is the constant of integration.
Substituting the values \[\dfrac{d}{dt}\left( \cos t \right)=\sin t\] and \[\int{{{e}^{t}}dt}={{e}^{t}}+c\] in equation (1), we will get
\[\begin{align}
  & I=\cos t\int{{{e}^{t}}dt-\int{\dfrac{d}{dt}\left( \cos t \right)\int{{{e}^{t}}dt}dt}} \\
 & ={{e}^{t}}\cos t+\int{{{e}^{t}}\sin tdt}+c.......(2)
\end{align}\]
Let us suppose that \[J\] denotes the integral \[\int{{{e}^{t}}\sin tdt}\].
That is, let \[J=\int{{{e}^{t}}\sin tdt}\].
Now using formula of integration by parts on both side of the above integral, we get
\[\begin{align}
  & I=\int{{{e}^{t}}\sin tdt} \\
 & =\sin t\int{{{e}^{t}}dt-\int{\dfrac{d}{dt}\left( \sin t \right)\int{{{e}^{t}}dt}dt}}................(3)
\end{align}\]
Now since we have \[\dfrac{d}{dt}\left( \sin t \right)=\cos t\] and \[\int{{{e}^{t}}dt}={{e}^{t}}+c\] where \[c\] is the constant of integration.
Substituting the values \[\dfrac{d}{dt}\left( \sin t \right)=\cos t\] and \[\int{{{e}^{t}}dt}={{e}^{t}}+c\] in equation (3), we will get
\[J={{e}^{t}}\sin t-\int{{{e}^{t}}\cos dt}+c........(4)\]
Thus on substituting integral \[I=\int{{{e}^{t}}\cos tdt}\] in equation (4), we have
\[J={{e}^{t}}\sin t-I+c\]
Now using \[J={{e}^{t}}\sin t-I+c\] in equation (2), we get
\[\begin{align}
  & I={{e}^{t}}\cos t+\int{{{e}^{t}}\sin tdt}+c \\
 & ={{e}^{t}}\cos t+{{e}^{t}}\sin t-I+c
\end{align}\]
Now we will take the integral \[I\] on one side,
\[2I={{e}^{t}}\cos t+{{e}^{t}}\sin t+c\]
On dividing the above equation by 2, we have
\[I=\dfrac{1}{2}{{e}^{t}}\cos t+\dfrac{1}{2}{{e}^{t}}\sin t+c\]
Now on substituting the value \[{{\log }_{e}}x=t\] in the above equation, we get
\[I=\dfrac{1}{2}{{e}^{{{\log }_{e}}x}}\cos \left( {{\log }_{e}}x \right)+\dfrac{1}{2}{{e}^{{{\log }_{e}}x}}\sin \left( {{\log }_{e}}x \right)+c\]
Since we know that \[{{e}^{{{\log }_{e}}x}}=x\], thus we have
\[I=\dfrac{1}{2}x\cos \left( {{\log }_{e}}x \right)+\dfrac{1}{2}x\sin \left( {{\log }_{e}}x \right)+c\]
Therefore we have
\[\begin{align}
  & \int{\cos \left( {{\log }_{e}}x \right)dx}=\dfrac{x}{2}\cos \left( {{\log }_{e}}x \right)+\dfrac{x}{2}\sin \left( {{\log }_{e}}x \right)+c \\
 & =\dfrac{x}{2}\left( \cos \left( {{\log }_{e}}x \right)+\sin \left( {{\log }_{e}}x \right) \right)+c
\end{align}\].

So, the correct answer is “Option B”.

Note: In this problem, we are evaluate the integral \[I=\int{\cos \left( {{\log }_{e}}x \right)dx}\] by substituting the value \[{{\log }_{e}}x=t\] in the integrant and then we are using integration by parts twice in the whole solution. Further do not forget to substitute the value \[{{\log }_{e}}x=t\] in the end to get the desired answer.