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

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

& I=\int{\cos \left( {{\log }_{e}}x \right)dx} \\\ & =\int{{{e}^{t}}\cos tdt} \end{aligned}$$ Since we know that by integration by parts we have $$\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{aligned} & 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{aligned}$$ Now since we have $$\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{aligned} & 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{aligned}$$ Let us suppose that $$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{aligned} & 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{aligned}$$ Now since we have $$\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{aligned} & I={{e}^{t}}\cos t+\int{{{e}^{t}}\sin tdt}+c \\\ & ={{e}^{t}}\cos t+{{e}^{t}}\sin t-I+c \end{aligned}$$ Now we will take the integral $$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{aligned} & \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{aligned}$$. **So, the correct answer is “Option B”.** **Note:** In this problem, we are evaluate the integral $$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.