Evaluate the following integrals: ( )dx - Vidyadip

Question

Evaluate the following integrals: $\int\cos(\log\text{x})\text{dx}$

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Answer

Let $\text{I}=\int\cos(\log\text{x})\text{dx}$
Let $\log\text{x}=\text{t}$
$\Rightarrow\text{x}=\text{e}^\text{t}$
$\Rightarrow\text{dx}=\text{e}^\text{t}\text{dt}$
$\text{I}=\int\text{e}^\text{t}\cos\text{(t)dt}$
Considering $\cos (t)$ as first function and $e^t$ as second function
$\text{I}=\cos\text{t}\text{e}^\text{t}-\int(-\sin\text{t})\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{t}\text{e}^\text{t}+\int\sin\text{t}\text{e}^\text{t}\text{dt}$
$\Rightarrow\text{I}=\cos\text{te}^\text{t}+\text{I}_1\ \dots(1)$
where $\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
$\text{I}_1=\int\text{e}^\text{t}\sin\text{t dt}$
Considering $\sin t$ as first function and $e^t$ as second function
$\text{I}_1=\sin\text{te}^\text{t}-\int\cos\text{te}^\text{t}\text{dt}$
$\Rightarrow\text{I}_1=\sin\text{te}^\text{t}-\text{I}\ \dots(2)$
From $(1) (2)$
$\text{I}=\cos\text{te}^\text{t}+\sin\text{te}^\text{t}-\text{I}$
$\Rightarrow2\text{I}=\text{e}^\text{t}(\sin\text{t}+\cos\text{t})$
$\Rightarrow\text{I}=\frac{\text{e}^\text{t}(\sin\text{t}+\cos\text{t})}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{e}^{\log\text{x}}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]}{2}+\text{C}$
$\Rightarrow\text{I}=\frac{\text{x}}{2}\big[\sin(\log\text{x})+\cos(\log\text{x})\big]+\text{C}$

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