Question

Evaluate $\int {\left[ {\sin \left( {\log x} \right) + \cos \left( {\log x} \right)} \right]} dx = $
A.$x\cos \left( {\log x} \right) + c$
B.$\cos \left( {\log x} \right) + c$
C.$x\sin \left( {\log x} \right) + c$
D.$\sin \left( {\log x} \right) + c$

Answer 1

Hint: In order to find the value of the integration of the given function, separately integrate the operands. We can see that there are two functions in each operand, so we would use the integration by parts method to integrate the functions and then substitute them in the original equation, solve them and get the results.
Formula used:
$\int {{e^t}dt} = {e^t}$
$\int {uvdx = u\int {vdx} - \int {u'\left( {\int {vdx} } \right)dx} } $
$\left( {\sin t} \right)' = \cos t$

Complete step-by-step answer:
We are given with an equation $\int {\left[ {\sin \left( {\log x} \right) + \cos \left( {\log x} \right)} \right]} dx$. Considering it to be $'I'$. Numerically written as:
$I = \int {\left[ {\sin \left( {\log x} \right) + \cos \left( {\log x} \right)} \right]} dx$
Taking the integration sign separately, we get:
$I = \int {\sin \left( {\log x} \right)} dx + \int {\cos \left( {\log x} \right)dx} $ ……..(1)
Taking the first operand to be ${I_1}$ and the other operand to be ${I_2}$, written as:
$I = {I_1} + {I_2}$
Solving the first operand ${I_1}$, we get:
${I_1} = \int {\sin \left( {\log x} \right)} dx$ ………(2)
Considering $\log x = t$.
Writing $\log x = t$ as $x = {e^t}$.
Differentiating both sides of $\log x = t$ with respect to x, we get:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{{dt}}{{dx}}$
$ \Rightarrow \dfrac{1}{x} = \dfrac{{dt}}{{dx}}$
$ \Rightarrow dx = xdt$
Substituting the value of $x = {e^t}$ in the above equation, we get:
$ \Rightarrow dx = {e^t}dt$
Substituting $dx = {e^t}dt$ and $\log x = t$ in equation 2, we get:
${I_1} = \int {\sin t.\left( {{e^t}} \right)dt} $
Since, we have two functions, so we are using integration by parts, which is as:
$\int {uvdx = u\int {vdx} - \int {u'\left( {\int {vdx} } \right)dx} } $
Comparing u, v with ${I_1}$, we get:
$
  u = \sin t \\
  v = {e^t} \;
 $
Substituting the values in the formula of parts, we get:
${I_1} = \int {\sin t.\left( {{e^t}} \right)dt = \sin t\int {{e^t}dt} - \int {\left( {\sin t} \right)'\left( {\int {{e^t}dt} } \right)dt} } $
Since, we know that $\int {{e^t}dt} = {e^t}$ and $\left( {\sin t} \right)' = \cos t$, so substituting the value in above equation, we get:
\[ \Rightarrow {I_1} = \int {\sin t.\left( {{e^t}} \right)dt = {e^t}\sin t - \int {{e^t}\cos tdt} } + c\]
Substituting the values ${e^t}$as $x$ and $t$ as $\left( {\log x} \right)$ on the right side of the above equation and we get:
$ \Rightarrow {I_1} = \int {\sin \left( {\log x} \right)dx = x\sin \left( {\log x} \right) - \int {x\cos \left( {\log x} \right)\dfrac{{dx}}{x}} } + c$
$ \Rightarrow {I_1} = \int {\sin \left( {\log x} \right)} dx = x\sin \left( {\log x} \right) - \int {\cos \left( {\log x} \right)dx} + c$
Substituting this equation in the equation 1, and we get:
$I = x\sin \left( {\log x} \right) - \int {\cos \left( {\log x} \right)dx} + \int {\cos \left( {\log x} \right)dx} + c$
And, we can see that the first and second operand can be cancelled, so we get:
$I = x\sin \left( {\log x} \right) + c$
$ \Rightarrow \int {\left[ {\sin \left( {\log x} \right) + \cos \left( {\log x} \right)} \right]} dx = x\sin \left( {\log x} \right) + c$
Hence, $\int {\left[ {\sin \left( {\log x} \right) + \cos \left( {\log x} \right)} \right]} dx = x\sin \left( {\log x} \right) + c$ which is the option 3.
So, the correct answer is “Option 3”.

Note: The term “Operands” used in the above statements refers to the equations, variables, or constants etc., present before and after an operator (+, -, *, /).
It is always recommended to solve the integration parts step by step in order to avoid any mistakes.