Question

How do you evaluate the integral of $\int{\dfrac{\cos \left( \text{ln x} \right)}{x}dx}$ ?

Answer 1

Hint: We are asked to evaluate the integral of $\dfrac{\cos \left( \text{ln }x \right)}{x}$ , to do so we will learn how integrals work, how the different functions are connected and how to integrate those functions. Then we see that in our given integral ln (x) and x are connected. We will use this to simplify and solve our given problem. We will use the formula that the integral of cos is sin and we will use substitution to solve our problem.

Complete step-by-step solution:
Before we start working on our problem we know about integral, integral or integration is just the opposite of the differentiation. In differentiation we usually split things to get something simpler while in integration we start with simplest (dx) and we build it to the biggest term.
Basic integration formula are $\int{{{x}^{n}}dn=\dfrac{{{x}^{n+1}}}{n+1}}$ , power is being increase by one and simultaneously divide by $n+1$ .
Now we are asked to evaluate integration $\int{\dfrac{\cos \left( \text{ln x} \right)}{x}dx}$
We can see that $\text{ln}\left( x \right)$ and x are connected to each other.
The derivative of $\ln \left( x \right)$ is $\dfrac{1}{x}$
So we will use this property of $\text{ln}\left( x \right)$ and we will solve our problem.
Now, we have to evaluate the integration of $\int{\dfrac{\cos \left( \text{ln x} \right)}{x}dx}$
We start our solution by substituting $\text{ln}\left( x \right)$ as ‘t’.
So $\ln \left( x \right)=t$
Now differentiating both sides, we get –
$\dfrac{1}{x}dx=dt$
So from here we get –
$dx=xdt$ ……………………… (1)
We will use $dx=xdt$ in our given integral.
So we get –
$\int{\dfrac{\cos \left( \text{ln }\left( \text{x} \right) \right)}{x}dx}=\int{\dfrac{\cos \left( t \right)}{x}xdt}$ (replaced ‘dx’ as ‘xdt’ and $\text{in}\left( x \right)$ as ‘t’)
So, simplifying we get –
$=\int{\cos \left( t \right)dt}$ (‘x’ cancel out)
Now as we know that integral of cos x is $\sin x+c$
So, integration of cos will be $\sin t+c$
We get –
\[=\sin \left( t \right)+c\]
Now replacing back the value of ‘t’ we put $t=\text{in}\left( x \right)$ back in our solution.
So we get –
$=\sin \left( \text{ln}\left( x \right) \right)+c$
Hence, we get – $\int{\dfrac{\cos \left( \text{ln x} \right)}{x}dx}=\sin \left( \ln \left( x \right) \right)+c$

Note: Remember that in case of indefinite integral, that is integral without limit it is necessary to replace or substitute back the original value of the function back which we replaced earlier.
While in case of definite integral no such thing is necessary as we change function, we also change the limit with it so we did not require changing back the function.
We simply apply the limit there but yes with change in function the limit must also be changed accordingly.