Mathematics

$$\int { \cos { \left( \log { x }  \right)  }  } dx=............\quad +\quad c$$


ANSWER

$$\dfrac { x }{ 2 } \left[ \cos { \left( \log { x } \right) } +\sin { \left( \log { x } \right) } \right] $$


SOLUTION

Consider the given integral.

$$I=\int{\cos \left( \log x \right)dx}$$

 

Let $$t=\log x$$

$$ \dfrac{dt}{dx}=\dfrac{1}{x} $$

$$ xdt=dx $$

 

Therefore,

$$ I=\int{{{e}^{t}}\cos tdt} $$

$$ I=\cos t{{e}^{t}}-\int{\left( -\sin t \right){{e}^{t}}}dt $$

$$ I={{e}^{t}}\cos t+\int{\left( \sin t \right){{e}^{t}}}dt $$

$$ I={{e}^{t}}\cos t+\sin t{{e}^{t}}-\int{\cos t{{e}^{t}}}dt $$

$$ I={{e}^{t}}\cos t+\sin t{{e}^{t}}-I $$

$$ 2I={{e}^{t}}\left( \cos t+\sin t \right)+C $$

$$ I=\dfrac{{{e}^{t}}}{2}\left( \cos t+\sin t \right)+C $$

 

On putting the value of $$'t'$$, we get

$$ I=\dfrac{{{e}^{\log x}}}{2}\left( \cos \left( \log x \right)+\sin \left( \log x \right) \right)+C $$

$$ I=\dfrac{x}{2}\left( \cos \left( \log x \right)+\sin \left( \log x \right) \right)+C $$

 

Hence, this is the answer.

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