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Find the value of by integrating \[\int {\sin \left( {\log x} \right)dx = } \]
A) \[\left( {\dfrac{x}{2}} \right)\left[ {\sin \left( {\log x} \right) - \cos \left( {\log x} \right)} \right]\]
B)\[\cos \left( {\log x} \right) - x\]
C)\[\int {\dfrac{{\left( {x - 1} \right){e^x}}}{{{{\left( {x + 1} \right)}^3}}}} \]
D)\[ - \cos \log x\]

Last updated date: 13th Jun 2024
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Hint:The above question is based on the concept of integration. Since it is an indefinite integral which has no upper and lower limits, we can apply integration properties by integrating it in parts and we can find the antiderivative of the above expression.

Complete step by step solution:
Integration is a way of finding the antiderivative of any function. It is the inverse of differentiation. It denotes the summation of discrete data. Calculation of small problems is an easy task but for adding big problems which include higher limits, integration method is used. The above given expression is an indefinite integral which means there are no upper or lower limits given.
The above expression after integrating should be in the below form.
\[\int {f\left( x \right) = F\left( x \right) + C} \]
where C is constant.
So, the given expression is \[\int {\sin \left( {\log x} \right)dx} \]
So we need to integrating it by parts by using the formula,
\[\int {uvdx = u\int {vdx - \int {\left\{ {\dfrac{d}{{dx}}\left( u \right)\int {vdx} } \right\}dx} } } \]
By applying it on the expression,
\Rightarrow \int {\sin \left( {\log x} \right)dx = \sin \left( {\log x} \right) \times x - \int {\left\{ {\cos
\left( {\log x} \right) \times \dfrac{1}{x} \times x} \right\}dx + C} } \\
\Rightarrow \int {\sin \left( {\log x} \right)dx = x\sin \left( {\log x} \right) - \int {\cos \left( {\log x}
\right)} } \\
Now the cosine function is again a composite function which means one function has another function inside it.Therefore we apply integration by parts on the last term.
= x\sin x\left( {\log x} \right) - \left[ {\cos \left( {\log x} \right)\int {dx} - \int {\left\{
{\dfrac{d}{{dx}}\cos \left( {\log x} \right)\int {dx} } \right\}} } \right] + C \\

= x\left\{ {\sin \left( {\log x} \right) - \cos \left( {\log x} \right)} \right\} - \int {\sin \left( {\log x}
\right)dx + C} \\
Now by shifting the sine function to the left-hand side we get,
\[ \Rightarrow 2\int {\sin \left( {\log x} \right)dx = \dfrac{x}{2}\left\{ {\sin \left( {\log x} \right) - \cos \left( {\log x} \right)} \right\} + \dfrac{c}{2}} \]

Note: An important thing to note is that integration of the number is the variable \[\int {1dx = x} \]. Since the number one can be written as \[{x^0}\] where the power is 0 .So when we integrate the power i.e., the power is added by the number one where 0+1=1.So it integrates to the variable \[{x^1}\].