Question

Evaluate the integral of $\int {{\text{cos}}\left( {{\text{ln x}}} \right){\text{dx}}} = $
A) $\dfrac{{\text{x}}}{{\text{2}}}\left( {{\text{cos ln x + sin ln x}}} \right)$
B) \[\dfrac{{\text{x}}}{{\text{2}}}\left[ {{\text{cos ln x - sin ln x}}} \right]\]
C) $\left[ {{\text{x cos ln x + sin ln x}}} \right]$
D) None of these

Answer 1

Hint: In this question, we have to evaluate the question to find the value of the given integral term. For that, we are going to solve that given integral term by using integration and differentiation methods. There are no limit values in the given integral term. So we only have to evaluate the values of the integral term.

Formula used: ${\text{log x = }}\dfrac{{\text{1}}}{{\text{x}}}$ where ${\text{x}}$ is a constant.
$\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{uv = u }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + v }}\dfrac{{{\text{du}}}}{{{\text{dx}}}}$
${\text{u dv = uv - }}\int {{\text{v du}}} $

Complete step-by-step answer:
Looking at $\int {{\text{cos}}\left( {{\text{ln x}}} \right){\text{dx}}} $, we realize that it can’t be integrated straight away. So we will use substitution to solve this question.
We would need ${\text{cos}}\left( {{\text{ln x}}} \right)\dfrac{{\text{1}}}{{\text{x}}}$ to integrate by substitution.
Running out of ideas, let’s try putting in the “missing” $\dfrac{{\text{1}}}{{\text{x}}}$ and an ${\text{x}}$ to make up for it and try integrating by parts.
$\Rightarrow$$\int {{\text{cos}}\left( {{\text{ln x}}} \right){\text{dx = }}\int {{\text{x cos}}\left( {{\text{ln x}}} \right)} {\text{. }}\dfrac{{\text{1}}}{{\text{x}}}{\text{dx}}} $---------(1)
Let ${\text{u = x}}$ and ${\text{dv = cos}}\left( {{\text{ln x}}} \right){\text{.}}\dfrac{{\text{1}}}{{\text{x}}}{\text{dx}}$
So, ${\text{du = dx}}$ and ${\text{v = sin}}\left( {{\text{ln x}}} \right)$
Substituting these values in (1), we get
$\Rightarrow$$\int {{\text{cos}}\left( {{\text{ln x}}} \right){\text{dx = x sin}}\left( {\ln {\text{x}}} \right) - \int {\sin \left( {{\text{ln x}}} \right)} {\text{ dx}}} $
$\Rightarrow${\[\int {{\text{sin}}\left( {{\text{ln x}}} \right){\text{dx}}} \] Here are going to insert ${\text{x }}{\text{. }}\dfrac{{\text{1}}}{{\text{x}}}$}
$\Rightarrow$$\int {{\text{cos}}\left( {{\text{ln x}}} \right){\text{dx = x sin}}\left( {\ln {\text{x}}} \right) - \int {{\text{x }}\sin \left( {{\text{ln x}}} \right){\text{ }}\dfrac{1}{{\text{x}}}} {\text{ dx}}} $
Now, we are going to integrate the term, we have
$\Rightarrow$$\int {{\text{x sin}}\left( {{\text{ln x}}} \right)\dfrac{{\text{1}}}{{\text{x}}}{\text{dx = }}\left[ {{\text{ - x cos}}\left( {{\text{ln x}}} \right){\text{ - }}\int {{\text{ - cos}}\left( {{\text{ln x}}} \right){\text{dx}}} } \right]} $
So we now have,
$\Rightarrow$$\int {\cos \left( {{\text{ln x}}} \right){\text{dx = x sin}}\left( {\ln {\text{x}}} \right){\text{ - }}\left[ {{\text{ - x cos}}\left( {{\text{ln x}}} \right){\text{ - }}\int {{\text{ - cos}}\left( {{\text{ln x}}} \right){\text{dx}}} } \right]} $
Simplifying we get,
$\Rightarrow$$\int {\cos \left( {{\text{ln x}}} \right){\text{dx = x sin}}\left( {\ln {\text{x}}} \right){\text{ + x cos}}\left( {{\text{ln x}}} \right) - \int {{\text{ cos}}\left( {{\text{ln x}}} \right){\text{dx}}} } $
Now let us take ${\text{I = }}\int {{\text{cos}}\left( {{\text{ln x}}} \right){\text{dx}}} $, then we have that
$\Rightarrow$\[{\text{I = x sin}}\left( {\ln {\text{x}}} \right) + {\text{x cos}}\left( {\ln {\text{x}}} \right){\text{ - I }}\]
By adding ${\text{I}}$ then we have that,
$\Rightarrow$\[{\text{2I = x sin}}\left( {\ln {\text{x}}} \right) + {\text{x cos}}\left( {\ln {\text{x}}} \right){\text{ }}\]
Rearranging the term for ${\text{I}}$,
$\Rightarrow$\[{\text{I = }}\dfrac{1}{2}\left[ {{\text{x sin}}\left( {\ln {\text{x}}} \right) + {\text{x cos}}\left( {\ln {\text{x}}} \right)} \right]{\text{ }}\]
Now substitute the values of ${\text{I}}$, we get
$\Rightarrow$\[\int {\cos \left( {\ln {\text{x}}} \right)} {\text{ dx = }}\dfrac{1}{2}\left[ {{\text{x sin}}\left( {\ln {\text{x}}} \right) + {\text{x cos}}\left( {\ln {\text{x}}} \right)} \right]{\text{ + c }}\]
Taking common \[{\text{x}}\] out from the brackets,
$\Rightarrow$\[\int {\cos \left( {\ln {\text{x}}} \right)} {\text{ dx = }}\dfrac{{\text{x}}}{2}\left[ {{\text{sin}}\left( {\ln {\text{x}}} \right) + {\text{cos}}\left( {\ln {\text{x}}} \right)} \right]{\text{ + c }}\]

Hence option A is the correct answer.

Note: We use definite integrals when the upper and lower limits of that function are given. We use indefinite integrals when no limits are given to a particular function. Integration is the inverse of differentiation.