Mathematics

Solve :
$$\int \cos (log x)dx$$ ?


SOLUTION
Given, $$\int { \cos { (\log { x } ) }  } dx$$
Let, $$\log { x } =t\\ \Rightarrow x={ e }^{ t }$$
and $$\dfrac { 1 }{ x } dx=dt\\ \Rightarrow dx=xdt\\ \Rightarrow dx={ e }^{ t }dt$$
So, we can write the integral as
$$\int { \cos { (\log { x } ) }  } dx\\ =\int { \cos { (t) } { e }^{ t } } dt\\ =\dfrac { 1 }{ 2 } \int { { e }^{ t } } (2\cos { t } )dt\\ =\dfrac { 1 }{ 2 } \int { { e }^{ t } } \left[ (\sin { t } +\cos { t } )+(\cos { t } -\sin { t } ) \right] dt$$
We know that, $$\int { { e }^{ t } } \left[ f\left( t \right) +f^{ ' }\left( t \right)  \right] dt={ e }^{ t }\left[ f\left( t \right)  \right] +C$$
Thus, $$f\left( t \right) =\sin { t } +\cos { t } $$ and $$f^{ ' }\left( t \right) =\cos { t } -\sin { t } $$
$$\therefore \dfrac { 1 }{ 2 } \int { { e }^{ t } } \left[ (\sin { t } +\cos { t } )+(\cos { t } -\sin { t } ) \right] dt\\ =(\dfrac { 1 }{ 2 } ){ e }^{ t }\left[ \sin { t } +\cos { t }  \right] +C\\ =\dfrac { 1 }{ 2 } x\left[ \sin { (\log { x } ) } +\cos { (\log { x } ) }  \right] +C$$
Hence, $$\int { \cos { (\log { x } ) }  } dx=\dfrac { 1 }{ 2 } x\left[ \sin { (\log { x } ) } +\cos { (\log { x } ) }  \right] +C.$$
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Subjective Medium Published on 17th 09, 2020
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