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Evaluate:

\(\smallint \frac{{{\rm{cos}}\left( {{\rm{ln}}\left( {\rm{x}} \right)} \right)}}{{\rm{x}}}{\rm{dx}}\)
1. sin(ln(x)) + C
2. cos(ln(x)) + C
3. \( - {\rm{}}\frac{{{\rm{cosx}}}}{{{{\rm{x}}^2}}} + {\rm{C}}\)
4. \(- \frac{{{\rm{sinx}}}}{{{{\rm{x}}^2}}} + {\rm{C}}\)
5. None of these

1 Answer

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Best answer
Correct Answer - Option 1 : sin(ln(x)) + C

Concept:

1. Integration by Substitution:

  • If the given integration is of the form \(\smallint {\rm{g}}\left( {{\rm{f}}\left( {\rm{x}} \right)} \right){\rm{f'}}\left( {\rm{x}} \right){\rm{dx}}\) where \({\rm{g}}\left( {\rm{x}} \right)\) and \({\rm{f}}\left( {\rm{x}} \right)\) are both differentiable functions then we substitute \({\rm{f}}\left( {\rm{x}} \right) = {\rm{u}}\) which implies that \({\rm{f'}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{du}}\).
  • Therefore, the integral becomes \(\smallint {\rm{g}}\left( {\rm{u}} \right){\rm{du}}\) which can be solved by general formulas.


Solution:

In the given problem substitute \(\ln \left( {\rm{x}} \right) = u\) therefore, \(\frac{{{\rm{dx}}}}{{\rm{x}}} = {\rm{du}}\).

The given integral becomes

\(\smallint \cos {\rm{udu}} = \sin {\rm{u}} + {\rm{C}}\)

Resubstitute \({\rm{u}} = \ln \left( {\rm{x}} \right)\).

\(\smallint \frac{{\cos \left( {\ln {\rm{x}}} \right)}}{{\rm{x}}}{\rm{dx}} = \sin \left( {\ln {\rm{x}}} \right) + {\rm{C}}\)

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